SlideShare a Scribd company logo
Rules For Differentiation
Rules For Differentiation(1) y c=
Rules For Differentiation(1) y c=
( )f x c=
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
( ) 0
lim
h
c c
f x
h→
−
′ =
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
( ) 0
lim
h
c c
f x
h→
−
′ =
0
0
lim
h h→
=
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
( ) 0
lim
h
c c
f x
h→
−
′ =
0
0
lim
h h→
=
0
lim0
0
h→
=
=
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
( ) 0
lim
h
c c
f x
h→
−
′ =
0
0
lim
h h→
=
0
lim0
0
h→
=
=
(2) y kx=
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
( ) 0
lim
h
c c
f x
h→
−
′ =
0
0
lim
h h→
=
0
lim0
0
h→
=
=
(2) y kx=
( )f x kx=
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
( ) 0
lim
h
c c
f x
h→
−
′ =
0
0
lim
h h→
=
0
lim0
0
h→
=
=
(2) y kx=
( )f x kx=
( ) ( )f x h k x h
kx kh
+ = +
= +
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
( ) 0
lim
h
c c
f x
h→
−
′ =
0
0
lim
h h→
=
0
lim0
0
h→
=
=
(2) y kx=
( )f x kx=
( ) ( )f x h k x h
kx kh
+ = +
= +
( ) 0
lim
h
kx kh kx
f x
h→
+ −
′ =
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
( ) 0
lim
h
c c
f x
h→
−
′ =
0
0
lim
h h→
=
0
lim0
0
h→
=
=
(2) y kx=
( )f x kx=
( ) ( )f x h k x h
kx kh
+ = +
= +
( ) 0
lim
h
kx kh kx
f x
h→
+ −
′ =
0
lim
h
kh
h→
=
Rules For Differentiation(1) y c=
( )f x c=
( )f x h c+ =
( ) 0
lim
h
c c
f x
h→
−
′ =
0
0
lim
h h→
=
0
lim0
0
h→
=
=
(2) y kx=
( )f x kx=
( ) ( )f x h k x h
kx kh
+ = +
= +
( ) 0
lim
h
kx kh kx
f x
h→
+ −
′ =
0
lim
h
kh
h→
=
0
lim
h
k
k
→
=
=
(3) n
y x=
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
( ) ( )3 3 2 2
a b a b a ab b− = − + +
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
( ) ( )3 3 2 2
a b a b a ab b− = − + +
( ) ( )4 4 3 2 2 3
a b a b a a b ab b− = − + + +
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
( ) ( )3 3 2 2
a b a b a ab b− = − + +
( ) ( )4 4 3 2 2 3
a b a b a a b ab b− = − + + +
( ) ( )1 2 3 2 2 3 2 1n n n n n n n n
a b a b a a b a b a b ab b− − − − − −
− = − + + + + + +
M
K
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
( ) ( )3 3 2 2
a b a b a ab b− = − + +
( ) ( )4 4 3 2 2 3
a b a b a a b ab b− = − + + +
( ) ( )1 2 3 2 2 3 2 1n n n n n n n n
a b a b a a b a b a b ab b− − − − − −
− = − + + + + + +
M
K
( ) n
f x x=
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
( ) ( )3 3 2 2
a b a b a ab b− = − + +
( ) ( )4 4 3 2 2 3
a b a b a a b ab b− = − + + +
( ) ( )1 2 3 2 2 3 2 1n n n n n n n n
a b a b a a b a b a b ab b− − − − − −
− = − + + + + + +
M
K
( ) n
f x x= ( ) ( )
n
f x h x h+ = +
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
( ) ( )3 3 2 2
a b a b a ab b− = − + +
( ) ( )4 4 3 2 2 3
a b a b a a b ab b− = − + + +
( ) ( )1 2 3 2 2 3 2 1n n n n n n n n
a b a b a a b a b a b ab b− − − − − −
− = − + + + + + +
M
K
( ) n
f x x= ( ) ( )
n
f x h x h+ = +
( )
( )
0
lim
n n
h
x h x
f x
h→
+ −
′ =
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
( ) ( )3 3 2 2
a b a b a ab b− = − + +
( ) ( )4 4 3 2 2 3
a b a b a a b ab b− = − + + +
( ) ( )1 2 3 2 2 3 2 1n n n n n n n n
a b a b a a b a b a b ab b− − − − − −
− = − + + + + + +
M
K
( ) n
f x x= ( ) ( )
n
f x h x h+ = +
( )
( )
0
lim
n n
h
x h x
f x
h→
+ −
′ =
( ) ( ) ( ) ( ){ }1 2 2 1
0
lim
n n n n
h
x h x x h x h x x h x x
h
− − − −
→
+ − + + + + + + +
=
K
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
( ) ( )3 3 2 2
a b a b a ab b− = − + +
( ) ( )4 4 3 2 2 3
a b a b a a b ab b− = − + + +
( ) ( )1 2 3 2 2 3 2 1n n n n n n n n
a b a b a a b a b a b ab b− − − − − −
− = − + + + + + +
M
K
( ) n
f x x= ( ) ( )
n
f x h x h+ = +
( )
( )
0
lim
n n
h
x h x
f x
h→
+ −
′ =
( ) ( ) ( ) ( ){ }1 2 2 1
0
lim
n n n n
h
x h x x h x h x x h x x
h
− − − −
→
+ − + + + + + + +
=
K
( ) ( ) ( ){ }1 2 2 1
0
lim
n n n n
h
h x h x h x x h x x
h
− − − −
→
+ + + + + + +
=
K
(3) n
y x=
( ) ( )2 2
a b a b a b− = − +
( ) ( )3 3 2 2
a b a b a ab b− = − + +
( ) ( )4 4 3 2 2 3
a b a b a a b ab b− = − + + +
( ) ( )1 2 3 2 2 3 2 1n n n n n n n n
a b a b a a b a b a b ab b− − − − − −
− = − + + + + + +
M
K
( ) n
f x x= ( ) ( )
n
f x h x h+ = +
( )
( )
0
lim
n n
h
x h x
f x
h→
+ −
′ =
( ) ( ) ( ) ( ){ }1 2 2 1
0
lim
n n n n
h
x h x x h x h x x h x x
h
− − − −
→
+ − + + + + + + +
=
K
( ) ( ) ( ){ }1 2 2 1
0
lim
n n n n
h
h x h x h x x h x x
h
− − − −
→
+ + + + + + +
=
K
( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
x h x h x x h x x
− − − −
→
= + + + + + + +K
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) 0
1 1
lim
h
x h xf x
h→
−
+′ =
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) 0
1 1
lim
h
x h xf x
h→
−
+′ =
( )0
lim
h
x x h
hx x h→
− −
=
+
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) 0
1 1
lim
h
x h xf x
h→
−
+′ =
( )0
lim
h
x x h
hx x h→
− −
=
+
( )0
lim
h
h
hx x h→
−
=
+
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) 0
1 1
lim
h
x h xf x
h→
−
+′ =
( )0
lim
h
x x h
hx x h→
− −
=
+
( )0
lim
h
h
hx x h→
−
=
+
( )0
1
lim
h x x h→
−
=
+
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) 0
1 1
lim
h
x h xf x
h→
−
+′ =
( )0
lim
h
x x h
hx x h→
− −
=
+
( )0
lim
h
h
hx x h→
−
=
+
( )0
1
lim
h x x h→
−
=
+
2
1
x
−
=
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) 0
1 1
lim
h
x h xf x
h→
−
+′ =
( )0
lim
h
x x h
hx x h→
− −
=
+
( )0
lim
h
h
hx x h→
−
=
+
( )0
1
lim
h x x h→
−
=
+
2
1
x
−
=
Note:
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) 0
1 1
lim
h
x h xf x
h→
−
+′ =
( )0
lim
h
x x h
hx x h→
− −
=
+
( )0
lim
h
h
hx x h→
−
=
+
( )0
1
lim
h x x h→
−
=
+
2
1
x
−
=
Note:
( ) 1
f x x−
=
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) 0
1 1
lim
h
x h xf x
h→
−
+′ =
( )0
lim
h
x x h
hx x h→
− −
=
+
( )0
lim
h
h
hx x h→
−
=
+
( )0
1
lim
h x x h→
−
=
+
2
1
x
−
=
Note:
( ) 1
f x x−
=
( ) 2
f x x−
′ = −
( ) ( ) ( ) ( )
1 2 2 1
0
lim
n n n n
h
f x x h x h x x h x x
− − − −
→
′ = + + + + + + +K
1 1 1 1n n n n
x x x x− − − −
= + + + +K
1n
nx −
=
1
(4) y
x
=
( )
1
f x
x
=
( )
1
f x h
x h
+ =
+
( ) 0
1 1
lim
h
x h xf x
h→
−
+′ =
( )0
lim
h
x x h
hx x h→
− −
=
+
( )0
lim
h
h
hx x h→
−
=
+
( )0
1
lim
h x x h→
−
=
+
2
1
x
−
=
Note:
( ) 1
f x x−
=
( ) 2
f x x−
′ = −
2
1
x
−
=
(5) y x=
(5) y x=
( )f x x=
(5) y x=
( )f x x=
( )f x h x h+ = +
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
( )0
lim
h
x h x
h x h x→
+ −
=
+ +
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
( )0
lim
h
x h x
h x h x→
+ −
=
+ +
( )0
lim
h
h
h x h x→
=
+ +
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
( )0
lim
h
x h x
h x h x→
+ −
=
+ +
( )0
lim
h
h
h x h x→
=
+ +
( )0
1
lim
h
x h x→
=
+ +
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
( )0
lim
h
x h x
h x h x→
+ −
=
+ +
( )0
lim
h
h
h x h x→
=
+ +
( )0
1
lim
h
x h x→
=
+ +
1
x x
=
+
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
( )0
lim
h
x h x
h x h x→
+ −
=
+ +
( )0
lim
h
h
h x h x→
=
+ +
( )0
1
lim
h
x h x→
=
+ +
1
x x
=
+
1
2 x
=
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
( )0
lim
h
x h x
h x h x→
+ −
=
+ +
( )0
lim
h
h
h x h x→
=
+ +
( )0
1
lim
h
x h x→
=
+ +
1
x x
=
+
1
2 x
=
Note:
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
( )0
lim
h
x h x
h x h x→
+ −
=
+ +
( )0
lim
h
h
h x h x→
=
+ +
( )0
1
lim
h
x h x→
=
+ +
1
x x
=
+
1
2 x
=
Note:
( )
1
2
f x x=
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
( )0
lim
h
x h x
h x h x→
+ −
=
+ +
( )0
lim
h
h
h x h x→
=
+ +
( )0
1
lim
h
x h x→
=
+ +
1
x x
=
+
1
2 x
=
Note:
( )
1
2
f x x=
( )
1
2
1
2
f x x
−
′ =
(5) y x=
( )f x x=
( )f x h x h+ = + ( ) 0
lim
h
x h x
f x
h→
+ −
′ =
x h x
x h x
+ +
×
+ +
( )0
lim
h
x h x
h x h x→
+ −
=
+ +
( )0
lim
h
h
h x h x→
=
+ +
( )0
1
lim
h
x h x→
=
+ +
1
x x
=
+
1
2 x
=
Note:
( )
1
2
f x x=
( )
1
2
1
2
f x x
−
′ =
1
2 x
=
( )e.g. 7i y =
( )e.g. 7i y =
0
dy
dx
=
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( ) 2
vii y x x=
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( ) 2
vii y x x=
5
2
x=
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( ) 2
vii y x x=
5
2
x=
3
2
5
2
dy
x
dx
=
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( ) 2
vii y x x=
5
2
x=
3
2
5
2
dy
x
dx
=
5
2
x x=
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( ) 2
vii y x x=
5
2
x=
3
2
5
2
dy
x
dx
=
5
2
x x=
( ) ( )
( )
3
If 3,
find 2
viii f x x
f
= −
′
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( ) 2
vii y x x=
5
2
x=
3
2
5
2
dy
x
dx
=
5
2
x x=
( ) ( )
( )
3
If 3,
find 2
viii f x x
f
= −
′
( ) 3
3f x x= −
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( ) 2
vii y x x=
5
2
x=
3
2
5
2
dy
x
dx
=
5
2
x x=
( ) ( )
( )
3
If 3,
find 2
viii f x x
f
= −
′
( ) 3
3f x x= −
( ) 2
3f x x′ =
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( ) 2
vii y x x=
5
2
x=
3
2
5
2
dy
x
dx
=
5
2
x x=
( ) ( )
( )
3
If 3,
find 2
viii f x x
f
= −
′
( ) 3
3f x x= −
( ) 2
3f x x′ =
( ) ( )
2
2 3 2f ′ =
( )e.g. 7i y =
0
dy
dx
=
( ) 37ii y x=
37
dy
dx
=
( ) 10
iii y x=
9
10
dy
x
dx
=
( ) 2
3 6 2iv y x x= + +
6 6
dy
x
dx
= +
( ) ( )
2
2 1v y x= +
2
4 4 1x x= + +
8 4
dy
x
dx
= +
( ) 2
1
3vi y x
x
= +
2
3x x−
= +
3
3 2
dy
x
dx
−
= −
3
2
3
x
= −
( ) 2
vii y x x=
5
2
x=
3
2
5
2
dy
x
dx
=
5
2
x x=
( ) ( )
( )
3
If 3,
find 2
viii f x x
f
= −
′
( ) 3
3f x x= −
( ) 2
3f x x′ =
( ) ( )
2
2 3 2f ′ =
12=
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
1 3 3y x− = −
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
1 3 3y x− = −
3 2 0x y− − =
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
1 3 3y x− = −
3 2 0x y− − =
( ) 3
Find the points on the curve 12 where the tangents
are horizontal
x y x x= −
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
1 3 3y x− = −
3 2 0x y− − =
( ) 3
Find the points on the curve 12 where the tangents
are horizontal
x y x x= −
3
12y x x= −
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
1 3 3y x− = −
3 2 0x y− − =
( ) 3
Find the points on the curve 12 where the tangents
are horizontal
x y x x= −
3
12y x x= −
2
3 12
dy
x
dx
= −
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
1 3 3y x− = −
3 2 0x y− − =
( ) 3
Find the points on the curve 12 where the tangents
are horizontal
x y x x= −
3
12y x x= −
2
3 12
dy
x
dx
= −
tangents are horizontal when 0
dy
dx
=
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
1 3 3y x− = −
3 2 0x y− − =
( ) 3
Find the points on the curve 12 where the tangents
are horizontal
x y x x= −
3
12y x x= −
2
3 12
dy
x
dx
= −
tangents are horizontal when 0
dy
dx
=
2
i.e. 3 12 0x − =
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
1 3 3y x− = −
3 2 0x y− − =
( ) 3
Find the points on the curve 12 where the tangents
are horizontal
x y x x= −
3
12y x x= −
2
3 12
dy
x
dx
= −
tangents are horizontal when 0
dy
dx
=
2
i.e. 3 12 0x − =
2
4
2
x
x
=
= ±
( )
( )
3 2
Find the equation of the tangent to the curve 5 6 2
at the point 1,1
xix y x x= − +
3 2
5 6 2y x x= − +
2
15 12
dy
x x
dx
= −
( ) ( )
2
when 1, 15 1 12 1
dy
x
dx
= = −
3=
required slope 3∴ =
( )1 3 1y x− = −
1 3 3y x− = −
3 2 0x y− − =
( ) 3
Find the points on the curve 12 where the tangents
are horizontal
x y x x= −
3
12y x x= −
2
3 12
dy
x
dx
= −
tangents are horizontal when 0
dy
dx
=
2
i.e. 3 12 0x − =
2
4
2
x
x
=
= ±
( ) ( )tangents are horizontal at 2,16 and 2, 16∴ − −
A normal is a line perpendicular to the tangent at the point of contact
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangent
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
( )
( )
2
Find the equation of the normal to the curve 4 3 2 at
the point 3,29
xi y x x= − +
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
( )
( )
2
Find the equation of the normal to the curve 4 3 2 at
the point 3,29
xi y x x= − +
2
4 3 2y x x= − +
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
( )
( )
2
Find the equation of the normal to the curve 4 3 2 at
the point 3,29
xi y x x= − +
2
4 3 2y x x= − +
8 3
dy
x
dx
= −
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
( )
( )
2
Find the equation of the normal to the curve 4 3 2 at
the point 3,29
xi y x x= − +
2
4 3 2y x x= − +
8 3
dy
x
dx
= −
( )when 3, 8 3 3
dy
x
dx
= = −
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
( )
( )
2
Find the equation of the normal to the curve 4 3 2 at
the point 3,29
xi y x x= − +
2
4 3 2y x x= − +
8 3
dy
x
dx
= −
( )when 3, 8 3 3
dy
x
dx
= = −
21=
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
( )
( )
2
Find the equation of the normal to the curve 4 3 2 at
the point 3,29
xi y x x= − +
2
4 3 2y x x= − +
8 3
dy
x
dx
= −
( )when 3, 8 3 3
dy
x
dx
= = −
21=
1
required slope
21
∴ = −
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
( )
( )
2
Find the equation of the normal to the curve 4 3 2 at
the point 3,29
xi y x x= − +
2
4 3 2y x x= − +
8 3
dy
x
dx
= −
( )when 3, 8 3 3
dy
x
dx
= = −
21=
1
required slope
21
∴ = −
( )
1
29 3
21
y x− = − −
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
( )
( )
2
Find the equation of the normal to the curve 4 3 2 at
the point 3,29
xi y x x= − +
2
4 3 2y x x= − +
8 3
dy
x
dx
= −
( )when 3, 8 3 3
dy
x
dx
= = −
21=
1
required slope
21
∴ = −
( )
1
29 3
21
y x− = − −
21 609 3y x− = − +
A normal is a line perpendicular to the tangent at the point of contact
y
x
( )y f x=
tangentnormal
( )
( )
2
Find the equation of the normal to the curve 4 3 2 at
the point 3,29
xi y x x= − +
2
4 3 2y x x= − +
8 3
dy
x
dx
= −
( )when 3, 8 3 3
dy
x
dx
= = −
21=
1
required slope
21
∴ = −
( )
1
29 3
21
y x− = − −
21 609 3y x− = − +
21 612 0x y+ − =
Exercise 7C; 1ace etc, 2ace etc, 3ace etc, 4bd, 5bdfh,
8bd, 9bd, 10ac, 12, 13b, 16, 21
Exercise 7D; 2ac, 3bd, 4ace, 6c, 7b, 11a, 13aei, 18, 22

More Related Content

What's hot

Integration formulas
Integration formulasIntegration formulas
Integration formulas
Muhammad Hassam
 
Cuaderno+de+integrales
Cuaderno+de+integralesCuaderno+de+integrales
Cuaderno+de+integralesjoseluisroyo
 
resposta do capitulo 15
resposta do capitulo 15resposta do capitulo 15
resposta do capitulo 15
silvio_sas
 
Solution Manual : Chapter - 02 Limits and Continuity
Solution Manual : Chapter - 02 Limits and ContinuitySolution Manual : Chapter - 02 Limits and Continuity
Solution Manual : Chapter - 02 Limits and Continuity
Hareem Aslam
 
Ejercicios varios de algebra widmar aguilar
Ejercicios varios de  algebra   widmar aguilarEjercicios varios de  algebra   widmar aguilar
Ejercicios varios de algebra widmar aguilar
Widmar Aguilar Gonzalez
 
Solution Manual : Chapter - 05 Integration
Solution Manual : Chapter - 05 IntegrationSolution Manual : Chapter - 05 Integration
Solution Manual : Chapter - 05 Integration
Hareem Aslam
 
solucionario de purcell 1
solucionario de purcell 1solucionario de purcell 1
solucionario de purcell 1
José Encalada
 

What's hot (12)

008 math a-net
008 math a-net008 math a-net
008 math a-net
 
Integration formulas
Integration formulasIntegration formulas
Integration formulas
 
Integral table
Integral tableIntegral table
Integral table
 
Basic m4-2-chapter1
Basic m4-2-chapter1Basic m4-2-chapter1
Basic m4-2-chapter1
 
Cuaderno+de+integrales
Cuaderno+de+integralesCuaderno+de+integrales
Cuaderno+de+integrales
 
Capitulo 7 Soluciones Purcell 9na Edicion
Capitulo 7 Soluciones Purcell 9na EdicionCapitulo 7 Soluciones Purcell 9na Edicion
Capitulo 7 Soluciones Purcell 9na Edicion
 
resposta do capitulo 15
resposta do capitulo 15resposta do capitulo 15
resposta do capitulo 15
 
Capitulo 5 Soluciones Purcell 9na Edicion
Capitulo 5 Soluciones Purcell 9na EdicionCapitulo 5 Soluciones Purcell 9na Edicion
Capitulo 5 Soluciones Purcell 9na Edicion
 
Solution Manual : Chapter - 02 Limits and Continuity
Solution Manual : Chapter - 02 Limits and ContinuitySolution Manual : Chapter - 02 Limits and Continuity
Solution Manual : Chapter - 02 Limits and Continuity
 
Ejercicios varios de algebra widmar aguilar
Ejercicios varios de  algebra   widmar aguilarEjercicios varios de  algebra   widmar aguilar
Ejercicios varios de algebra widmar aguilar
 
Solution Manual : Chapter - 05 Integration
Solution Manual : Chapter - 05 IntegrationSolution Manual : Chapter - 05 Integration
Solution Manual : Chapter - 05 Integration
 
solucionario de purcell 1
solucionario de purcell 1solucionario de purcell 1
solucionario de purcell 1
 

Viewers also liked

Product and Quotient Rule Show
Product and Quotient Rule ShowProduct and Quotient Rule Show
Product and Quotient Rule Showguest4fb2de
 
cobb douglas production function
cobb douglas production functioncobb douglas production function
cobb douglas production function
Lakshmanamoorthi Moorthi
 
Lesson 10: The Chain Rule (Section 41 slides)
Lesson 10: The Chain Rule (Section 41 slides)Lesson 10: The Chain Rule (Section 41 slides)
Lesson 10: The Chain Rule (Section 41 slides)
Matthew Leingang
 
11 x1 t09 05 product rule (2012)
11 x1 t09 05 product rule (2012)11 x1 t09 05 product rule (2012)
11 x1 t09 05 product rule (2012)Nigel Simmons
 
2.7 chain rule short cuts
2.7 chain rule short cuts2.7 chain rule short cuts
2.7 chain rule short cutsmath265
 
EE&FA - Cobb-Douglas Production Function - FINAL YEAR CS/IT - SRI SAIRAM INST...
EE&FA - Cobb-Douglas Production Function - FINAL YEAR CS/IT - SRI SAIRAM INST...EE&FA - Cobb-Douglas Production Function - FINAL YEAR CS/IT - SRI SAIRAM INST...
EE&FA - Cobb-Douglas Production Function - FINAL YEAR CS/IT - SRI SAIRAM INST...
SRI SAIRAM INSTITUTE OF TECHNOLOGY, CHENNAI
 
Lesson 23: The Chain Rule
Lesson 23: The Chain RuleLesson 23: The Chain Rule
Lesson 23: The Chain Rule
Matthew Leingang
 
Core 3 The Product Rule
Core 3 The Product RuleCore 3 The Product Rule
Core 3 The Product Rule
davidmiles100
 
CHAIN RULE AND IMPLICIT FUNCTION
CHAIN RULE AND IMPLICIT FUNCTIONCHAIN RULE AND IMPLICIT FUNCTION
CHAIN RULE AND IMPLICIT FUNCTION
Nikhil Pandit
 
Cobb-douglas production function
Cobb-douglas production functionCobb-douglas production function
Cobb-douglas production function
Suniya Sheikh
 
Cobb douglas
Cobb douglasCobb douglas
Cobb douglascr_mau
 
Lecture 8 derivative rules
Lecture 8   derivative rulesLecture 8   derivative rules
Lecture 8 derivative rules
njit-ronbrown
 
Core 3 The Chain Rule
Core 3 The Chain RuleCore 3 The Chain Rule
Core 3 The Chain Rule
davidmiles100
 
Cobb Douglas production function
Cobb Douglas production functionCobb Douglas production function
Cobb Douglas production function
Suniya Sheikh
 
Core 3 The Quotient Rule
Core 3 The Quotient RuleCore 3 The Quotient Rule
Core 3 The Quotient Rule
davidmiles100
 
The Building Block of Calculus - Chapter 4 The Product Rule and Quotient Rule
The Building Block of Calculus - Chapter 4 The Product Rule and Quotient RuleThe Building Block of Calculus - Chapter 4 The Product Rule and Quotient Rule
The Building Block of Calculus - Chapter 4 The Product Rule and Quotient Rule
Tenri Ashari Wanahari
 
Presentation on CVP Analysis, Break Even Point & Applications of Marginal Cos...
Presentation on CVP Analysis, Break Even Point & Applications of Marginal Cos...Presentation on CVP Analysis, Break Even Point & Applications of Marginal Cos...
Presentation on CVP Analysis, Break Even Point & Applications of Marginal Cos...Leena Kakkar
 
Cobb douglas production function
Cobb douglas production functionCobb douglas production function
Cobb douglas production functionMahdi Mesbahi
 
Cobb-Douglas Production Function
Cobb-Douglas  Production  FunctionCobb-Douglas  Production  Function
Cobb-Douglas Production FunctionNabduan Duangmanee
 

Viewers also liked (20)

Chain Rule
Chain RuleChain Rule
Chain Rule
 
Product and Quotient Rule Show
Product and Quotient Rule ShowProduct and Quotient Rule Show
Product and Quotient Rule Show
 
cobb douglas production function
cobb douglas production functioncobb douglas production function
cobb douglas production function
 
Lesson 10: The Chain Rule (Section 41 slides)
Lesson 10: The Chain Rule (Section 41 slides)Lesson 10: The Chain Rule (Section 41 slides)
Lesson 10: The Chain Rule (Section 41 slides)
 
11 x1 t09 05 product rule (2012)
11 x1 t09 05 product rule (2012)11 x1 t09 05 product rule (2012)
11 x1 t09 05 product rule (2012)
 
2.7 chain rule short cuts
2.7 chain rule short cuts2.7 chain rule short cuts
2.7 chain rule short cuts
 
EE&FA - Cobb-Douglas Production Function - FINAL YEAR CS/IT - SRI SAIRAM INST...
EE&FA - Cobb-Douglas Production Function - FINAL YEAR CS/IT - SRI SAIRAM INST...EE&FA - Cobb-Douglas Production Function - FINAL YEAR CS/IT - SRI SAIRAM INST...
EE&FA - Cobb-Douglas Production Function - FINAL YEAR CS/IT - SRI SAIRAM INST...
 
Lesson 23: The Chain Rule
Lesson 23: The Chain RuleLesson 23: The Chain Rule
Lesson 23: The Chain Rule
 
Core 3 The Product Rule
Core 3 The Product RuleCore 3 The Product Rule
Core 3 The Product Rule
 
CHAIN RULE AND IMPLICIT FUNCTION
CHAIN RULE AND IMPLICIT FUNCTIONCHAIN RULE AND IMPLICIT FUNCTION
CHAIN RULE AND IMPLICIT FUNCTION
 
Cobb-douglas production function
Cobb-douglas production functionCobb-douglas production function
Cobb-douglas production function
 
Cobb douglas
Cobb douglasCobb douglas
Cobb douglas
 
Lecture 8 derivative rules
Lecture 8   derivative rulesLecture 8   derivative rules
Lecture 8 derivative rules
 
Core 3 The Chain Rule
Core 3 The Chain RuleCore 3 The Chain Rule
Core 3 The Chain Rule
 
Cobb Douglas production function
Cobb Douglas production functionCobb Douglas production function
Cobb Douglas production function
 
Core 3 The Quotient Rule
Core 3 The Quotient RuleCore 3 The Quotient Rule
Core 3 The Quotient Rule
 
The Building Block of Calculus - Chapter 4 The Product Rule and Quotient Rule
The Building Block of Calculus - Chapter 4 The Product Rule and Quotient RuleThe Building Block of Calculus - Chapter 4 The Product Rule and Quotient Rule
The Building Block of Calculus - Chapter 4 The Product Rule and Quotient Rule
 
Presentation on CVP Analysis, Break Even Point & Applications of Marginal Cos...
Presentation on CVP Analysis, Break Even Point & Applications of Marginal Cos...Presentation on CVP Analysis, Break Even Point & Applications of Marginal Cos...
Presentation on CVP Analysis, Break Even Point & Applications of Marginal Cos...
 
Cobb douglas production function
Cobb douglas production functionCobb douglas production function
Cobb douglas production function
 
Cobb-Douglas Production Function
Cobb-Douglas  Production  FunctionCobb-Douglas  Production  Function
Cobb-Douglas Production Function
 

Similar to 11 x1 t09 03 rules for differentiation (2013)

Integration formulas
Integration formulasIntegration formulas
Integration formulas
Sri Chakra Kumar
 
01 derivadas
01   derivadas01   derivadas
01 derivadasklorofila
 
comp diff
comp diffcomp diff
comp diffdianenz
 
51542 0131469657 ism-1
51542 0131469657 ism-151542 0131469657 ism-1
51542 0131469657 ism-1
Joel Rodriguez Llamuca
 
1. limits
1. limits1. limits
Formulario de Calculo Diferencial-Integral
Formulario de Calculo Diferencial-IntegralFormulario de Calculo Diferencial-Integral
Formulario de Calculo Diferencial-Integral
Erick Chevez
 
2. the derivative
2. the derivative2. the derivative
2. the derivative
Esteban Alzate Pérez
 
51543 0131469657 ism-2
51543 0131469657 ism-251543 0131469657 ism-2
51543 0131469657 ism-2Carlos Fuentes
 
Formulario cálculo
Formulario cálculoFormulario cálculo
Formulario cálculo
Man50035
 
Ejerciciosderivadasresueltos
EjerciciosderivadasresueltosEjerciciosderivadasresueltos
Ejerciciosderivadasresueltosbellidomates
 
Formulario oficial-calculo
Formulario oficial-calculoFormulario oficial-calculo
Formulario oficial-calculo
Favian Flores
 
51542 0131469657 ism-1
51542 0131469657 ism-151542 0131469657 ism-1
51542 0131469657 ism-1
Ani_Agustina
 
jhkl,l.มือครูคณิตศาสตร์พื้นฐาน ม.4 สสวท เล่ม 2fuyhfg
jhkl,l.มือครูคณิตศาสตร์พื้นฐาน ม.4 สสวท เล่ม 2fuyhfgjhkl,l.มือครูคณิตศาสตร์พื้นฐาน ม.4 สสวท เล่ม 2fuyhfg
jhkl,l.มือครูคณิตศาสตร์พื้นฐาน ม.4 สสวท เล่ม 2fuyhfg
Tonn Za
 
Clase 16 dsp
Clase 16 dspClase 16 dsp
Clase 16 dsp
Marcelo Valdiviezo
 
Formulario derivadas e integrales
Formulario derivadas e integralesFormulario derivadas e integrales
Formulario derivadas e integralesGeovanny Jiménez
 
Formulario
FormularioFormulario
Formulario
Jonnathan Cespedes
 
Formulario
FormularioFormulario
Formulario
Genaro Coronel
 

Similar to 11 x1 t09 03 rules for differentiation (2013) (20)

Integration formulas
Integration formulasIntegration formulas
Integration formulas
 
01 derivadas
01   derivadas01   derivadas
01 derivadas
 
comp diff
comp diffcomp diff
comp diff
 
51542 0131469657 ism-1
51542 0131469657 ism-151542 0131469657 ism-1
51542 0131469657 ism-1
 
1. limits
1. limits1. limits
1. limits
 
Formulario de Calculo Diferencial-Integral
Formulario de Calculo Diferencial-IntegralFormulario de Calculo Diferencial-Integral
Formulario de Calculo Diferencial-Integral
 
2. the derivative
2. the derivative2. the derivative
2. the derivative
 
51543 0131469657 ism-2
51543 0131469657 ism-251543 0131469657 ism-2
51543 0131469657 ism-2
 
Capitulo 2 Soluciones Purcell 9na Edicion
Capitulo 2 Soluciones Purcell 9na EdicionCapitulo 2 Soluciones Purcell 9na Edicion
Capitulo 2 Soluciones Purcell 9na Edicion
 
Formulario calculo
Formulario calculoFormulario calculo
Formulario calculo
 
Formulario cálculo
Formulario cálculoFormulario cálculo
Formulario cálculo
 
Ejerciciosderivadasresueltos
EjerciciosderivadasresueltosEjerciciosderivadasresueltos
Ejerciciosderivadasresueltos
 
Formulario oficial-calculo
Formulario oficial-calculoFormulario oficial-calculo
Formulario oficial-calculo
 
51542 0131469657 ism-1
51542 0131469657 ism-151542 0131469657 ism-1
51542 0131469657 ism-1
 
Puteri de matrici
Puteri de matriciPuteri de matrici
Puteri de matrici
 
jhkl,l.มือครูคณิตศาสตร์พื้นฐาน ม.4 สสวท เล่ม 2fuyhfg
jhkl,l.มือครูคณิตศาสตร์พื้นฐาน ม.4 สสวท เล่ม 2fuyhfgjhkl,l.มือครูคณิตศาสตร์พื้นฐาน ม.4 สสวท เล่ม 2fuyhfg
jhkl,l.มือครูคณิตศาสตร์พื้นฐาน ม.4 สสวท เล่ม 2fuyhfg
 
Clase 16 dsp
Clase 16 dspClase 16 dsp
Clase 16 dsp
 
Formulario derivadas e integrales
Formulario derivadas e integralesFormulario derivadas e integrales
Formulario derivadas e integrales
 
Formulario
FormularioFormulario
Formulario
 
Formulario
FormularioFormulario
Formulario
 

More from Nigel Simmons

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATE
Nigel Simmons
 
Goodbye slideshare
Goodbye slideshareGoodbye slideshare
Goodbye slideshare
Nigel Simmons
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)Nigel Simmons
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)Nigel Simmons
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)Nigel Simmons
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)Nigel Simmons
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)Nigel Simmons
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)Nigel Simmons
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)Nigel Simmons
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)Nigel Simmons
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)Nigel Simmons
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)Nigel Simmons
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)Nigel Simmons
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)Nigel Simmons
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)Nigel Simmons
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)Nigel Simmons
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)Nigel Simmons
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)Nigel Simmons
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)Nigel Simmons
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)Nigel Simmons
 

More from Nigel Simmons (20)

Goodbye slideshare UPDATE
Goodbye slideshare UPDATEGoodbye slideshare UPDATE
Goodbye slideshare UPDATE
 
Goodbye slideshare
Goodbye slideshareGoodbye slideshare
Goodbye slideshare
 
12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)12 x1 t02 02 integrating exponentials (2014)
12 x1 t02 02 integrating exponentials (2014)
 
11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)11 x1 t01 03 factorising (2014)
11 x1 t01 03 factorising (2014)
 
11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)11 x1 t01 02 binomial products (2014)
11 x1 t01 02 binomial products (2014)
 
12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)12 x1 t02 01 differentiating exponentials (2014)
12 x1 t02 01 differentiating exponentials (2014)
 
11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)11 x1 t01 01 algebra & indices (2014)
11 x1 t01 01 algebra & indices (2014)
 
12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)12 x1 t01 03 integrating derivative on function (2013)
12 x1 t01 03 integrating derivative on function (2013)
 
12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)12 x1 t01 02 differentiating logs (2013)
12 x1 t01 02 differentiating logs (2013)
 
12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)12 x1 t01 01 log laws (2013)
12 x1 t01 01 log laws (2013)
 
X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)X2 t02 04 forming polynomials (2013)
X2 t02 04 forming polynomials (2013)
 
X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)X2 t02 03 roots & coefficients (2013)
X2 t02 03 roots & coefficients (2013)
 
X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)X2 t02 02 multiple roots (2013)
X2 t02 02 multiple roots (2013)
 
X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)X2 t02 01 factorising complex expressions (2013)
X2 t02 01 factorising complex expressions (2013)
 
11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)11 x1 t16 07 approximations (2013)
11 x1 t16 07 approximations (2013)
 
11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)11 x1 t16 06 derivative times function (2013)
11 x1 t16 06 derivative times function (2013)
 
11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)11 x1 t16 05 volumes (2013)
11 x1 t16 05 volumes (2013)
 
11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)11 x1 t16 04 areas (2013)
11 x1 t16 04 areas (2013)
 
11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)11 x1 t16 03 indefinite integral (2013)
11 x1 t16 03 indefinite integral (2013)
 
11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)11 x1 t16 02 definite integral (2013)
11 x1 t16 02 definite integral (2013)
 

Recently uploaded

Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
Welcome to TechSoup   New Member Orientation and Q&A (May 2024).pdfWelcome to TechSoup   New Member Orientation and Q&A (May 2024).pdf
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
TechSoup
 
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptxChapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptx
Mohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
 
The Roman Empire A Historical Colossus.pdf
The Roman Empire A Historical Colossus.pdfThe Roman Empire A Historical Colossus.pdf
The Roman Empire A Historical Colossus.pdf
kaushalkr1407
 
Digital Tools and AI for Teaching Learning and Research
Digital Tools and AI for Teaching Learning and ResearchDigital Tools and AI for Teaching Learning and Research
Digital Tools and AI for Teaching Learning and Research
Vikramjit Singh
 
"Protectable subject matters, Protection in biotechnology, Protection of othe...
"Protectable subject matters, Protection in biotechnology, Protection of othe..."Protectable subject matters, Protection in biotechnology, Protection of othe...
"Protectable subject matters, Protection in biotechnology, Protection of othe...
SACHIN R KONDAGURI
 
CACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdfCACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdf
camakaiclarkmusic
 
Operation Blue Star - Saka Neela Tara
Operation Blue Star   -  Saka Neela TaraOperation Blue Star   -  Saka Neela Tara
Operation Blue Star - Saka Neela Tara
Balvir Singh
 
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
MysoreMuleSoftMeetup
 
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
siemaillard
 
A Strategic Approach: GenAI in Education
A Strategic Approach: GenAI in EducationA Strategic Approach: GenAI in Education
A Strategic Approach: GenAI in Education
Peter Windle
 
Lapbook sobre os Regimes Totalitários.pdf
Lapbook sobre os Regimes Totalitários.pdfLapbook sobre os Regimes Totalitários.pdf
Lapbook sobre os Regimes Totalitários.pdf
Jean Carlos Nunes Paixão
 
Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk - How can education support child empowerment.pptxFrancesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk - How can education support child empowerment.pptx
EduSkills OECD
 
Model Attribute Check Company Auto Property
Model Attribute  Check Company Auto PropertyModel Attribute  Check Company Auto Property
Model Attribute Check Company Auto Property
Celine George
 
Phrasal Verbs.XXXXXXXXXXXXXXXXXXXXXXXXXX
Phrasal Verbs.XXXXXXXXXXXXXXXXXXXXXXXXXXPhrasal Verbs.XXXXXXXXXXXXXXXXXXXXXXXXXX
Phrasal Verbs.XXXXXXXXXXXXXXXXXXXXXXXXXX
MIRIAMSALINAS13
 
Language Across the Curriculm LAC B.Ed.
Language Across the  Curriculm LAC B.Ed.Language Across the  Curriculm LAC B.Ed.
Language Across the Curriculm LAC B.Ed.
Atul Kumar Singh
 
The approach at University of Liverpool.pptx
The approach at University of Liverpool.pptxThe approach at University of Liverpool.pptx
The approach at University of Liverpool.pptx
Jisc
 
Acetabularia Information For Class 9 .docx
Acetabularia Information For Class 9  .docxAcetabularia Information For Class 9  .docx
Acetabularia Information For Class 9 .docx
vaibhavrinwa19
 
2024.06.01 Introducing a competency framework for languag learning materials ...
2024.06.01 Introducing a competency framework for languag learning materials ...2024.06.01 Introducing a competency framework for languag learning materials ...
2024.06.01 Introducing a competency framework for languag learning materials ...
Sandy Millin
 
Sha'Carri Richardson Presentation 202345
Sha'Carri Richardson Presentation 202345Sha'Carri Richardson Presentation 202345
Sha'Carri Richardson Presentation 202345
beazzy04
 
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCECLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
BhavyaRajput3
 

Recently uploaded (20)

Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
Welcome to TechSoup   New Member Orientation and Q&A (May 2024).pdfWelcome to TechSoup   New Member Orientation and Q&A (May 2024).pdf
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
 
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptxChapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptx
 
The Roman Empire A Historical Colossus.pdf
The Roman Empire A Historical Colossus.pdfThe Roman Empire A Historical Colossus.pdf
The Roman Empire A Historical Colossus.pdf
 
Digital Tools and AI for Teaching Learning and Research
Digital Tools and AI for Teaching Learning and ResearchDigital Tools and AI for Teaching Learning and Research
Digital Tools and AI for Teaching Learning and Research
 
"Protectable subject matters, Protection in biotechnology, Protection of othe...
"Protectable subject matters, Protection in biotechnology, Protection of othe..."Protectable subject matters, Protection in biotechnology, Protection of othe...
"Protectable subject matters, Protection in biotechnology, Protection of othe...
 
CACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdfCACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdf
 
Operation Blue Star - Saka Neela Tara
Operation Blue Star   -  Saka Neela TaraOperation Blue Star   -  Saka Neela Tara
Operation Blue Star - Saka Neela Tara
 
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
 
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
 
A Strategic Approach: GenAI in Education
A Strategic Approach: GenAI in EducationA Strategic Approach: GenAI in Education
A Strategic Approach: GenAI in Education
 
Lapbook sobre os Regimes Totalitários.pdf
Lapbook sobre os Regimes Totalitários.pdfLapbook sobre os Regimes Totalitários.pdf
Lapbook sobre os Regimes Totalitários.pdf
 
Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk - How can education support child empowerment.pptxFrancesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk - How can education support child empowerment.pptx
 
Model Attribute Check Company Auto Property
Model Attribute  Check Company Auto PropertyModel Attribute  Check Company Auto Property
Model Attribute Check Company Auto Property
 
Phrasal Verbs.XXXXXXXXXXXXXXXXXXXXXXXXXX
Phrasal Verbs.XXXXXXXXXXXXXXXXXXXXXXXXXXPhrasal Verbs.XXXXXXXXXXXXXXXXXXXXXXXXXX
Phrasal Verbs.XXXXXXXXXXXXXXXXXXXXXXXXXX
 
Language Across the Curriculm LAC B.Ed.
Language Across the  Curriculm LAC B.Ed.Language Across the  Curriculm LAC B.Ed.
Language Across the Curriculm LAC B.Ed.
 
The approach at University of Liverpool.pptx
The approach at University of Liverpool.pptxThe approach at University of Liverpool.pptx
The approach at University of Liverpool.pptx
 
Acetabularia Information For Class 9 .docx
Acetabularia Information For Class 9  .docxAcetabularia Information For Class 9  .docx
Acetabularia Information For Class 9 .docx
 
2024.06.01 Introducing a competency framework for languag learning materials ...
2024.06.01 Introducing a competency framework for languag learning materials ...2024.06.01 Introducing a competency framework for languag learning materials ...
2024.06.01 Introducing a competency framework for languag learning materials ...
 
Sha'Carri Richardson Presentation 202345
Sha'Carri Richardson Presentation 202345Sha'Carri Richardson Presentation 202345
Sha'Carri Richardson Presentation 202345
 
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCECLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
CLASS 11 CBSE B.St Project AIDS TO TRADE - INSURANCE
 

11 x1 t09 03 rules for differentiation (2013)

  • 4. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ =
  • 5. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ = ( ) 0 lim h c c f x h→ − ′ =
  • 6. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ = ( ) 0 lim h c c f x h→ − ′ = 0 0 lim h h→ =
  • 7. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ = ( ) 0 lim h c c f x h→ − ′ = 0 0 lim h h→ = 0 lim0 0 h→ = =
  • 8. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ = ( ) 0 lim h c c f x h→ − ′ = 0 0 lim h h→ = 0 lim0 0 h→ = = (2) y kx=
  • 9. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ = ( ) 0 lim h c c f x h→ − ′ = 0 0 lim h h→ = 0 lim0 0 h→ = = (2) y kx= ( )f x kx=
  • 10. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ = ( ) 0 lim h c c f x h→ − ′ = 0 0 lim h h→ = 0 lim0 0 h→ = = (2) y kx= ( )f x kx= ( ) ( )f x h k x h kx kh + = + = +
  • 11. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ = ( ) 0 lim h c c f x h→ − ′ = 0 0 lim h h→ = 0 lim0 0 h→ = = (2) y kx= ( )f x kx= ( ) ( )f x h k x h kx kh + = + = + ( ) 0 lim h kx kh kx f x h→ + − ′ =
  • 12. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ = ( ) 0 lim h c c f x h→ − ′ = 0 0 lim h h→ = 0 lim0 0 h→ = = (2) y kx= ( )f x kx= ( ) ( )f x h k x h kx kh + = + = + ( ) 0 lim h kx kh kx f x h→ + − ′ = 0 lim h kh h→ =
  • 13. Rules For Differentiation(1) y c= ( )f x c= ( )f x h c+ = ( ) 0 lim h c c f x h→ − ′ = 0 0 lim h h→ = 0 lim0 0 h→ = = (2) y kx= ( )f x kx= ( ) ( )f x h k x h kx kh + = + = + ( ) 0 lim h kx kh kx f x h→ + − ′ = 0 lim h kh h→ = 0 lim h k k → = =
  • 15. (3) n y x= ( ) ( )2 2 a b a b a b− = − +
  • 16. (3) n y x= ( ) ( )2 2 a b a b a b− = − + ( ) ( )3 3 2 2 a b a b a ab b− = − + +
  • 17. (3) n y x= ( ) ( )2 2 a b a b a b− = − + ( ) ( )3 3 2 2 a b a b a ab b− = − + + ( ) ( )4 4 3 2 2 3 a b a b a a b ab b− = − + + +
  • 18. (3) n y x= ( ) ( )2 2 a b a b a b− = − + ( ) ( )3 3 2 2 a b a b a ab b− = − + + ( ) ( )4 4 3 2 2 3 a b a b a a b ab b− = − + + + ( ) ( )1 2 3 2 2 3 2 1n n n n n n n n a b a b a a b a b a b ab b− − − − − − − = − + + + + + + M K
  • 19. (3) n y x= ( ) ( )2 2 a b a b a b− = − + ( ) ( )3 3 2 2 a b a b a ab b− = − + + ( ) ( )4 4 3 2 2 3 a b a b a a b ab b− = − + + + ( ) ( )1 2 3 2 2 3 2 1n n n n n n n n a b a b a a b a b a b ab b− − − − − − − = − + + + + + + M K ( ) n f x x=
  • 20. (3) n y x= ( ) ( )2 2 a b a b a b− = − + ( ) ( )3 3 2 2 a b a b a ab b− = − + + ( ) ( )4 4 3 2 2 3 a b a b a a b ab b− = − + + + ( ) ( )1 2 3 2 2 3 2 1n n n n n n n n a b a b a a b a b a b ab b− − − − − − − = − + + + + + + M K ( ) n f x x= ( ) ( ) n f x h x h+ = +
  • 21. (3) n y x= ( ) ( )2 2 a b a b a b− = − + ( ) ( )3 3 2 2 a b a b a ab b− = − + + ( ) ( )4 4 3 2 2 3 a b a b a a b ab b− = − + + + ( ) ( )1 2 3 2 2 3 2 1n n n n n n n n a b a b a a b a b a b ab b− − − − − − − = − + + + + + + M K ( ) n f x x= ( ) ( ) n f x h x h+ = + ( ) ( ) 0 lim n n h x h x f x h→ + − ′ =
  • 22. (3) n y x= ( ) ( )2 2 a b a b a b− = − + ( ) ( )3 3 2 2 a b a b a ab b− = − + + ( ) ( )4 4 3 2 2 3 a b a b a a b ab b− = − + + + ( ) ( )1 2 3 2 2 3 2 1n n n n n n n n a b a b a a b a b a b ab b− − − − − − − = − + + + + + + M K ( ) n f x x= ( ) ( ) n f x h x h+ = + ( ) ( ) 0 lim n n h x h x f x h→ + − ′ = ( ) ( ) ( ) ( ){ }1 2 2 1 0 lim n n n n h x h x x h x h x x h x x h − − − − → + − + + + + + + + = K
  • 23. (3) n y x= ( ) ( )2 2 a b a b a b− = − + ( ) ( )3 3 2 2 a b a b a ab b− = − + + ( ) ( )4 4 3 2 2 3 a b a b a a b ab b− = − + + + ( ) ( )1 2 3 2 2 3 2 1n n n n n n n n a b a b a a b a b a b ab b− − − − − − − = − + + + + + + M K ( ) n f x x= ( ) ( ) n f x h x h+ = + ( ) ( ) 0 lim n n h x h x f x h→ + − ′ = ( ) ( ) ( ) ( ){ }1 2 2 1 0 lim n n n n h x h x x h x h x x h x x h − − − − → + − + + + + + + + = K ( ) ( ) ( ){ }1 2 2 1 0 lim n n n n h h x h x h x x h x x h − − − − → + + + + + + + = K
  • 24. (3) n y x= ( ) ( )2 2 a b a b a b− = − + ( ) ( )3 3 2 2 a b a b a ab b− = − + + ( ) ( )4 4 3 2 2 3 a b a b a a b ab b− = − + + + ( ) ( )1 2 3 2 2 3 2 1n n n n n n n n a b a b a a b a b a b ab b− − − − − − − = − + + + + + + M K ( ) n f x x= ( ) ( ) n f x h x h+ = + ( ) ( ) 0 lim n n h x h x f x h→ + − ′ = ( ) ( ) ( ) ( ){ }1 2 2 1 0 lim n n n n h x h x x h x h x x h x x h − − − − → + − + + + + + + + = K ( ) ( ) ( ){ }1 2 2 1 0 lim n n n n h h x h x h x x h x x h − − − − → + + + + + + + = K ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h x h x h x x h x x − − − − → = + + + + + + +K
  • 25. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K
  • 26. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K
  • 27. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − =
  • 28. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x =
  • 29. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x =
  • 30. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = +
  • 31. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = + ( ) 0 1 1 lim h x h xf x h→ − +′ =
  • 32. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = + ( ) 0 1 1 lim h x h xf x h→ − +′ = ( )0 lim h x x h hx x h→ − − = +
  • 33. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = + ( ) 0 1 1 lim h x h xf x h→ − +′ = ( )0 lim h x x h hx x h→ − − = + ( )0 lim h h hx x h→ − = +
  • 34. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = + ( ) 0 1 1 lim h x h xf x h→ − +′ = ( )0 lim h x x h hx x h→ − − = + ( )0 lim h h hx x h→ − = + ( )0 1 lim h x x h→ − = +
  • 35. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = + ( ) 0 1 1 lim h x h xf x h→ − +′ = ( )0 lim h x x h hx x h→ − − = + ( )0 lim h h hx x h→ − = + ( )0 1 lim h x x h→ − = + 2 1 x − =
  • 36. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = + ( ) 0 1 1 lim h x h xf x h→ − +′ = ( )0 lim h x x h hx x h→ − − = + ( )0 lim h h hx x h→ − = + ( )0 1 lim h x x h→ − = + 2 1 x − = Note:
  • 37. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = + ( ) 0 1 1 lim h x h xf x h→ − +′ = ( )0 lim h x x h hx x h→ − − = + ( )0 lim h h hx x h→ − = + ( )0 1 lim h x x h→ − = + 2 1 x − = Note: ( ) 1 f x x− =
  • 38. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = + ( ) 0 1 1 lim h x h xf x h→ − +′ = ( )0 lim h x x h hx x h→ − − = + ( )0 lim h h hx x h→ − = + ( )0 1 lim h x x h→ − = + 2 1 x − = Note: ( ) 1 f x x− = ( ) 2 f x x− ′ = −
  • 39. ( ) ( ) ( ) ( ) 1 2 2 1 0 lim n n n n h f x x h x h x x h x x − − − − → ′ = + + + + + + +K 1 1 1 1n n n n x x x x− − − − = + + + +K 1n nx − = 1 (4) y x = ( ) 1 f x x = ( ) 1 f x h x h + = + ( ) 0 1 1 lim h x h xf x h→ − +′ = ( )0 lim h x x h hx x h→ − − = + ( )0 lim h h hx x h→ − = + ( )0 1 lim h x x h→ − = + 2 1 x − = Note: ( ) 1 f x x− = ( ) 2 f x x− ′ = − 2 1 x − =
  • 41. (5) y x= ( )f x x=
  • 42. (5) y x= ( )f x x= ( )f x h x h+ = +
  • 43. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ =
  • 44. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + +
  • 45. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + + ( )0 lim h x h x h x h x→ + − = + +
  • 46. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + + ( )0 lim h x h x h x h x→ + − = + + ( )0 lim h h h x h x→ = + +
  • 47. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + + ( )0 lim h x h x h x h x→ + − = + + ( )0 lim h h h x h x→ = + + ( )0 1 lim h x h x→ = + +
  • 48. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + + ( )0 lim h x h x h x h x→ + − = + + ( )0 lim h h h x h x→ = + + ( )0 1 lim h x h x→ = + + 1 x x = +
  • 49. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + + ( )0 lim h x h x h x h x→ + − = + + ( )0 lim h h h x h x→ = + + ( )0 1 lim h x h x→ = + + 1 x x = + 1 2 x =
  • 50. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + + ( )0 lim h x h x h x h x→ + − = + + ( )0 lim h h h x h x→ = + + ( )0 1 lim h x h x→ = + + 1 x x = + 1 2 x = Note:
  • 51. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + + ( )0 lim h x h x h x h x→ + − = + + ( )0 lim h h h x h x→ = + + ( )0 1 lim h x h x→ = + + 1 x x = + 1 2 x = Note: ( ) 1 2 f x x=
  • 52. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + + ( )0 lim h x h x h x h x→ + − = + + ( )0 lim h h h x h x→ = + + ( )0 1 lim h x h x→ = + + 1 x x = + 1 2 x = Note: ( ) 1 2 f x x= ( ) 1 2 1 2 f x x − ′ =
  • 53. (5) y x= ( )f x x= ( )f x h x h+ = + ( ) 0 lim h x h x f x h→ + − ′ = x h x x h x + + × + + ( )0 lim h x h x h x h x→ + − = + + ( )0 lim h h h x h x→ = + + ( )0 1 lim h x h x→ = + + 1 x x = + 1 2 x = Note: ( ) 1 2 f x x= ( ) 1 2 1 2 f x x − ′ = 1 2 x =
  • 54. ( )e.g. 7i y =
  • 55. ( )e.g. 7i y = 0 dy dx =
  • 56. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x=
  • 57. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx =
  • 58. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x=
  • 59. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx =
  • 60. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + +
  • 61. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = +
  • 62. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= +
  • 63. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + +
  • 64. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = +
  • 65. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = +
  • 66. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = +
  • 67. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = −
  • 68. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = −
  • 69. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = − ( ) 2 vii y x x=
  • 70. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = − ( ) 2 vii y x x= 5 2 x=
  • 71. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = − ( ) 2 vii y x x= 5 2 x= 3 2 5 2 dy x dx =
  • 72. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = − ( ) 2 vii y x x= 5 2 x= 3 2 5 2 dy x dx = 5 2 x x=
  • 73. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = − ( ) 2 vii y x x= 5 2 x= 3 2 5 2 dy x dx = 5 2 x x= ( ) ( ) ( ) 3 If 3, find 2 viii f x x f = − ′
  • 74. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = − ( ) 2 vii y x x= 5 2 x= 3 2 5 2 dy x dx = 5 2 x x= ( ) ( ) ( ) 3 If 3, find 2 viii f x x f = − ′ ( ) 3 3f x x= −
  • 75. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = − ( ) 2 vii y x x= 5 2 x= 3 2 5 2 dy x dx = 5 2 x x= ( ) ( ) ( ) 3 If 3, find 2 viii f x x f = − ′ ( ) 3 3f x x= − ( ) 2 3f x x′ =
  • 76. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = − ( ) 2 vii y x x= 5 2 x= 3 2 5 2 dy x dx = 5 2 x x= ( ) ( ) ( ) 3 If 3, find 2 viii f x x f = − ′ ( ) 3 3f x x= − ( ) 2 3f x x′ = ( ) ( ) 2 2 3 2f ′ =
  • 77. ( )e.g. 7i y = 0 dy dx = ( ) 37ii y x= 37 dy dx = ( ) 10 iii y x= 9 10 dy x dx = ( ) 2 3 6 2iv y x x= + + 6 6 dy x dx = + ( ) ( ) 2 2 1v y x= + 2 4 4 1x x= + + 8 4 dy x dx = + ( ) 2 1 3vi y x x = + 2 3x x− = + 3 3 2 dy x dx − = − 3 2 3 x = − ( ) 2 vii y x x= 5 2 x= 3 2 5 2 dy x dx = 5 2 x x= ( ) ( ) ( ) 3 If 3, find 2 viii f x x f = − ′ ( ) 3 3f x x= − ( ) 2 3f x x′ = ( ) ( ) 2 2 3 2f ′ = 12=
  • 78. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − +
  • 79. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − +
  • 80. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = −
  • 81. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = −
  • 82. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3=
  • 83. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ =
  • 84. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = −
  • 85. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = − 1 3 3y x− = −
  • 86. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = − 1 3 3y x− = − 3 2 0x y− − =
  • 87. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = − 1 3 3y x− = − 3 2 0x y− − = ( ) 3 Find the points on the curve 12 where the tangents are horizontal x y x x= −
  • 88. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = − 1 3 3y x− = − 3 2 0x y− − = ( ) 3 Find the points on the curve 12 where the tangents are horizontal x y x x= − 3 12y x x= −
  • 89. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = − 1 3 3y x− = − 3 2 0x y− − = ( ) 3 Find the points on the curve 12 where the tangents are horizontal x y x x= − 3 12y x x= − 2 3 12 dy x dx = −
  • 90. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = − 1 3 3y x− = − 3 2 0x y− − = ( ) 3 Find the points on the curve 12 where the tangents are horizontal x y x x= − 3 12y x x= − 2 3 12 dy x dx = − tangents are horizontal when 0 dy dx =
  • 91. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = − 1 3 3y x− = − 3 2 0x y− − = ( ) 3 Find the points on the curve 12 where the tangents are horizontal x y x x= − 3 12y x x= − 2 3 12 dy x dx = − tangents are horizontal when 0 dy dx = 2 i.e. 3 12 0x − =
  • 92. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = − 1 3 3y x− = − 3 2 0x y− − = ( ) 3 Find the points on the curve 12 where the tangents are horizontal x y x x= − 3 12y x x= − 2 3 12 dy x dx = − tangents are horizontal when 0 dy dx = 2 i.e. 3 12 0x − = 2 4 2 x x = = ±
  • 93. ( ) ( ) 3 2 Find the equation of the tangent to the curve 5 6 2 at the point 1,1 xix y x x= − + 3 2 5 6 2y x x= − + 2 15 12 dy x x dx = − ( ) ( ) 2 when 1, 15 1 12 1 dy x dx = = − 3= required slope 3∴ = ( )1 3 1y x− = − 1 3 3y x− = − 3 2 0x y− − = ( ) 3 Find the points on the curve 12 where the tangents are horizontal x y x x= − 3 12y x x= − 2 3 12 dy x dx = − tangents are horizontal when 0 dy dx = 2 i.e. 3 12 0x − = 2 4 2 x x = = ± ( ) ( )tangents are horizontal at 2,16 and 2, 16∴ − −
  • 94. A normal is a line perpendicular to the tangent at the point of contact
  • 95. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x=
  • 96. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangent
  • 97. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal
  • 98. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal ( ) ( ) 2 Find the equation of the normal to the curve 4 3 2 at the point 3,29 xi y x x= − +
  • 99. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal ( ) ( ) 2 Find the equation of the normal to the curve 4 3 2 at the point 3,29 xi y x x= − + 2 4 3 2y x x= − +
  • 100. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal ( ) ( ) 2 Find the equation of the normal to the curve 4 3 2 at the point 3,29 xi y x x= − + 2 4 3 2y x x= − + 8 3 dy x dx = −
  • 101. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal ( ) ( ) 2 Find the equation of the normal to the curve 4 3 2 at the point 3,29 xi y x x= − + 2 4 3 2y x x= − + 8 3 dy x dx = − ( )when 3, 8 3 3 dy x dx = = −
  • 102. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal ( ) ( ) 2 Find the equation of the normal to the curve 4 3 2 at the point 3,29 xi y x x= − + 2 4 3 2y x x= − + 8 3 dy x dx = − ( )when 3, 8 3 3 dy x dx = = − 21=
  • 103. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal ( ) ( ) 2 Find the equation of the normal to the curve 4 3 2 at the point 3,29 xi y x x= − + 2 4 3 2y x x= − + 8 3 dy x dx = − ( )when 3, 8 3 3 dy x dx = = − 21= 1 required slope 21 ∴ = −
  • 104. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal ( ) ( ) 2 Find the equation of the normal to the curve 4 3 2 at the point 3,29 xi y x x= − + 2 4 3 2y x x= − + 8 3 dy x dx = − ( )when 3, 8 3 3 dy x dx = = − 21= 1 required slope 21 ∴ = − ( ) 1 29 3 21 y x− = − −
  • 105. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal ( ) ( ) 2 Find the equation of the normal to the curve 4 3 2 at the point 3,29 xi y x x= − + 2 4 3 2y x x= − + 8 3 dy x dx = − ( )when 3, 8 3 3 dy x dx = = − 21= 1 required slope 21 ∴ = − ( ) 1 29 3 21 y x− = − − 21 609 3y x− = − +
  • 106. A normal is a line perpendicular to the tangent at the point of contact y x ( )y f x= tangentnormal ( ) ( ) 2 Find the equation of the normal to the curve 4 3 2 at the point 3,29 xi y x x= − + 2 4 3 2y x x= − + 8 3 dy x dx = − ( )when 3, 8 3 3 dy x dx = = − 21= 1 required slope 21 ∴ = − ( ) 1 29 3 21 y x− = − − 21 609 3y x− = − + 21 612 0x y+ − =
  • 107. Exercise 7C; 1ace etc, 2ace etc, 3ace etc, 4bd, 5bdfh, 8bd, 9bd, 10ac, 12, 13b, 16, 21 Exercise 7D; 2ac, 3bd, 4ace, 6c, 7b, 11a, 13aei, 18, 22