chain rule calcalus

1. THE CHAIN RULE
(CASE 1)

2. 1. THE CHAIN RULE (CASE 1) Suppose that
z=f (x, y) is a differentiable function of x and y,
where x=g (t) and y=h (t) and are both
differentiable functions of t. Then z is a
differentiable function of t and
dt
dy
y
f
dt
dx
x
f
dt
dz
∂
∂
+
∂
∂
=
dt
dy
y
z
dt
dx
x
z
dt
dz
∂
∂
+
∂
∂
=

3. Question No. 1.
Use THE CHAIN RULE to find the derivative of
w = xy with respect to t along the path x = cost
and y = sint. What is the derivative’s value at t
= Pie/2.
).........(i
dt
dy
y
w
dt
dx
x
w
dt
dw
∂
∂
+
∂
∂
=
Solution:
We Use THE CHAIN RULE to find the
derivative of w = xy with respect to t along the
path x = cost and y = sint as follows:
x
y
xy
dy
dw
andy
x
xy
dx
dw
=
∂
∂
==
∂
∂
=
)()(

4. t
dt
td
dt
dy
andt
dt
td
dt
dx
cos
sin
sin
cos
==−==
Solution continued……….
ttt
tttt
txty
dt
dy
y
w
dt
dx
x
w
dt
dw
ieqnfromThen
2cossincos
).(coscos)sin.(sin
)(cos)sin(
)(
22
=−=
+−=
+−=
∂
∂
+
∂
∂
=
1cos
2
.2cos
2/
−==
=
=
π
π
πt
dt
dw
partFurther

5. 2. THE CHAIN RULE (CASE 2) Suppose that
w=f (x, y, z) is a differentiable function of x, y
and z and are all differentiable functions of t.
Then w is a differentiable function of t and
dt
dz
z
f
dt
dy
y
f
dt
dx
x
f
dt
dw
∂
∂
+
∂
∂
+
∂
∂
=
dt
dz
z
w
dt
dy
y
w
dt
dx
x
w
dt
dw
∂
∂
+
∂
∂
+
∂
∂
=

6. Question No. 2.
Use THE CHAIN RULE to find the derivative of
w = xy + z with respect to t along the path x =
cost , y = sint and z = t. What is the derivative’s
value at t = 0.
).........(i
dt
dz
z
w
dt
dy
y
w
dt
dx
x
w
dt
dw
∂
∂
+
∂
∂
+
∂
∂
=
Solution:
We Use THE CHAIN RULE to find the
derivative of w = xy + z with respect to t along
the path x = cost, y = sint and z = t as follows:
1
)()(
,
)(
=
∂
+∂
==
∂
+∂
==
∂
+∂
=
z
zxy
dz
dw
andx
y
zxy
dy
dw
andy
x
zxy
dx
dw

7. 1cos
sin
,sin
cos
====−==
dt
dt
dt
dz
andt
dt
td
dt
dy
t
dt
td
dt
dx
Solution continued……….
ttt
tttt
txty
dt
dz
z
w
dt
dy
y
w
dt
dx
x
w
dt
dw
ieqnfromThen
2cos11sincos
1).(coscos)sin.(sin
1.1)(cos)sin(
)(
22
+=+−=
++−=
++−=
∂
∂
+
∂
∂
+
∂
∂
=
211
0.2cos1
0
=+=
+=
=t
dt
dw
partFurther

8. 3. THE CHAIN RULE (CASE 3) Suppose that z=f
(x, y) is a differentiable function of x and y, where
x=g (s, t) and y=h (s, t) are differentiable
functions of s and t. Then
ds
dy
y
z
ds
dx
x
z
dx
dz
∂
∂
+
∂
∂
=
dt
dy
y
z
dt
dx
x
z
dt
dz
∂
∂
+
∂
∂
=

9. 4. THE CHAIN RULE (GENERAL VERSION)
Suppose that u is a differentiable function of the n
variables x1, x2,‧‧‧,xn and each xj is a differentiable
function of the m variables t1, t2,‧‧‧,tm Then u is a
function of t1, t2,‧‧‧, tm and
for each i=1,2,‧‧‧,m.
i
n
niii t
x
x
u
‧‧‧
dt
x
x
u
dt
dx
x
u
t
u
∂
∂
∂
∂
++
∂
∂
∂
+
∂
∂
=
∂
∂ 2
2
1
1

11. F (x, y)=0. Since both x and y are functions of
x, we obtain
But dx /dx=1, so if ∂F/∂y≠0 we solve for
dy/dx and obtain
0=
∂
∂
+
∂
∂
dx
dy
y
F
dx
dx
x
F
y
x
F
F
y
F
x
F
dx
dy
−=
∂
∂
∂
∂
−=

12. F (x, y, z)=0
But and
so this equation becomes
If ∂F/∂z≠0 ,we solve for ∂z/∂x and obtain the
first formula in Equations 7. The formula for
∂z/∂y is obtained in a similar manner.
0=
∂
∂
∂
∂
+
∂
∂
+
∂
∂
x
z
z
F
dx
dy
y
F
dx
dx
x
F
1)( =
∂
∂
x
x
1)( =
∂
∂
y
x
0=
∂
∂
∂
∂
+
∂
∂
x
z
z
F
x
F
z
F
x
F
dx
dz
∂
∂
∂
∂
−=
z
F
y
F
dy
dz
∂
∂
∂
∂
−=

15. 1. DEFINITION A function of two variables
has a local maximum at (a, b) if f (x, y) ≤ f
(a, b) when (x, y) is near (a, b). [This means
that
f (x, y) ≤ f (a, b) for all points (x, y) in some
disk with center (a, b).] The number f (a, b) is
called a local maximum value. If f (x, y) ≥ f
(a, b) when (x, y) is near (a, b), then f (a, b) is
a local minimum value.

16. 2. THEOREM If f has a local maximum or
minimum at (a, b) and the first order partial
derivatives of f exist there, then fx(a, b)=1 and
fy(a, b)=0.

17. A point (a, b) is called a critical point (or
stationary point) of f if fx (a, b)=0 and fy (a,
b)=0, or if one of these partial derivatives does
not exist.

19. 3. SECOND DERIVATIVES TEST Suppose the
second partial derivatives of f are continuous
on a disk with center (a, b) , and suppose that
fx (a, b) and fy (a, b)=0 [that is, (a, b) is a critical
point of f]. Let
(a)If D>0 and fxx (a, b)>0 , then f (a, b) is a local
minimum.
(b)If D>0 and fxx (a, b)<0, then f (a, b) is a local
maximum.
(c) If D<0, then f (a, b) is not a local maximum
or minimum.
2
)],([),(),(),( bafbafbafbaDD xyyyxx −==

20. NOTE 1 In case (c) the point (a, b) is called a
saddle point of f and the graph of f crosses its
tangent plane at (a, b).
NOTE 2 If D=0, the test gives no information:
f could have a local maximum or local
minimum at (a, b), or (a, b) could be a saddle
point of f.
NOTE 3 To remember the formula for D it’s
helpful to write it as a determinant:
2
)( xyyyxx
yyyx
xyxy
fff
ff
ff
D −==

21. 1444
+−+= xyyxz

24. 4. EXTREME VALUE THEOREM FOR
FUNCTIONS OF TWO VARIABLES If f is
continuous on a closed, bounded set D in R2
,
then f attains an absolute maximum value
f(x1,y1) and an absolute minimum value f(x2,y2)
at some points (x1,y1) and (x2,y2) in D.

25. 5. To find the absolute maximum and minimum
values of a continuous function f on a closed,
bounded set D:
1. Find the values of f at the critical points of in D.
2. Find the extreme values of f on the boundary of D.
3. The largest of the values from steps 1 and 2 is the
absolute maximum value; the smallest of these
values is the absolute minimum value.