This document discusses rules for computing derivatives of functions. It begins by listing existing derivative rules and defining notation. It then derives and presents rules for the derivatives of trigonometric functions like sine, cosine, tangent, cotangent, secant and cosecant. An example problem demonstrates finding the derivative of the tangent function using previous rules.

1. More Computations of Derivatives

2. More Computations of Derivatives
We list the derivative rules we have so far.

3. More Computations of Derivatives
We list the derivative rules we have so far.
We assume that the derivatives exist in all the
theorems below and the prime ( )' operation is
differentiation with respect to x.

4. More Computations of Derivatives
We list the derivative rules we have so far.
We assume that the derivatives exist in all the
theorems below and the prime ( )' operation is
differentiation with respect to x.
The ± and Constant–Multiple Derivative Rules

5. More Computations of Derivatives
We list the derivative rules we have so far.
We assume that the derivatives exist in all the
theorems below and the prime ( )' operation is
differentiation with respect to x.
The ± and Constant–Multiple Derivative Rules
Let f(x) and g(x) be two functions then
i. (f(x)±g(x)) ' = f '(x)±g '(x)
ii. (cf(x)) ' = c*f '(x) where c is a constant.

6. More Computations of Derivatives
We list the derivative rules we have so far.
We assume that the derivatives exist in all the
theorems below and the prime ( )' operation is
differentiation with respect to x.
The ± and Constant–Multiple Derivative Rules
Let f(x) and g(x) be two functions then
i. (f(x)±g(x)) ' = f '(x)±g '(x)
ii. (cf(x)) ' = c*f '(x) where c is a constant.
The Product and Quotient Rules of Derivatives

7. More Computations of Derivatives
We list the derivative rules we have so far.
We assume that the derivatives exist in all the
theorems below and the prime ( )' operation is
differentiation with respect to x.
The ± and Constant–Multiple Derivative Rules
Let f(x) and g(x) be two functions then
i. (f(x)±g(x)) ' = f '(x)±g '(x)
ii. (cf(x)) ' = c*f '(x) where c is a constant.
The Product and Quotient Rules of Derivatives
Write f for f(x) and g for g(x) then
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
The Product Rule
The Quotient Rule

8. More Computations of Derivatives
Derivatives of Constants and Power Functions
i. If f(x) = c, a constant function, then f '(x) = (c)' = 0.
ii. If f(x) = xP, then f '(x) = PxP–1 for a constant P ≠ 0.

9. More Computations of Derivatives
Derivatives of Constants and Power Functions
i. If f(x) = c, a constant function, then f '(x) = (c)' = 0.
ii. If f(x) = xP, then f '(x) = PxP–1 for a constant P ≠ 0.
From the above rules and the derivatives of the
monomials we are able to easily calculate the
derivative of polynomials and rational functions.

10. More Computations of Derivatives
Derivatives of Constants and Power Functions
i. If f(x) = c, a constant function, then f '(x) = (c)' = 0.
ii. If f(x) = xP, then f '(x) = PxP–1 for a constant P ≠ 0.
From the above rules and the derivatives of the
monomials we are able to easily calculate the
derivative of polynomials and rational functions.
But in order to take the derivatives of all elementary
functions, we also need to know the derivatives of the
other basic functions i.e. functions of the trig. family,
the log and exponential functions.

11. More Computations of Derivatives
Derivatives of Constants and Power Functions
i. If f(x) = c, a constant function, then f '(x) = (c)' = 0.
ii. If f(x) = xP, then f '(x) = PxP–1 for a constant P ≠ 0.
From the above rules and the derivatives of the
monomials we are able to easily calculate the
derivative of polynomials and rational functions.
But in order to take the derivatives of all elementary
functions, we also need to know the derivatives of the
other basic functions i.e. functions of the trig. family,
the log and exponential functions.
More importantly we need to know how the derivative
of the composition of two functions such as sin(x2) may
be expressed as the derivative of each component.

12. More Computations of Derivatives
Derivatives of Sine and Cosine
i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
ii. cos'(x) = –sin(x)
d cos(x)
dx = –sin(x)

13. More Computations of Derivatives
Derivatives of Sine and Cosine
i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
ii. cos'(x) = –sin(x)
d cos(x)
dx = –sin(x)
In the following verification of i, we write s(x) for sin(x)
and c(x) for cos(x).

14. More Computations of Derivatives
Derivatives of Sine and Cosine
i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
ii. cos'(x) = –sin(x)
d cos(x)
dx = –sin(x)
In the following verification of i, we write s(x) for sin(x)
and c(x) for cos(x). We need the following limit–results.

15. More Computations of Derivatives
Derivatives of Sine and Cosine
i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
ii. cos'(x) = –sin(x)
d cos(x)
dx = –sin(x)
In the following verification of i, we write s(x) for sin(x)
and c(x) for cos(x). We need the following limit–results.
lim
sin(h)
h
= 1
h→0
lim
cos(h) – 1
h
= 0
h→0
(HW)

16. More Computations of Derivatives
Derivatives of Sine and Cosine
i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
ii. cos'(x) = –sin(x)
d cos(x)
dx = –sin(x)
In the following verification of i, we write s(x) for sin(x)
and c(x) for cos(x). We need the following limit–results.
lim
sin(h)
h
= 1
h→0
cos(h) – 1
Expand the difference quotient of sin(x) by the Sum of
Angles Formula for sine.
s(x + h) – s(x)
h =
lim
h
= 0
h→0
(HW)

17. More Computations of Derivatives
Derivatives of Sine and Cosine
i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
ii. cos'(x) = –sin(x)
d cos(x)
dx = –sin(x)
In the following verification of i, we write s(x) for sin(x)
and c(x) for cos(x). We need the following limit–results.
lim
sin(h)
h
= 1
h→0
cos(h) – 1
Expand the difference quotient of sin(x) by the Sum of
Angles Formula for sine.
s(x + h) – s(x)
h =
s(x)c(h) + c(x)s(h) – s(x)
h
lim
h
= 0
h→0
(HW)

18. More Computations of Derivatives
Derivatives of Sine and Cosine
i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
ii. cos'(x) = –sin(x)
d cos(x)
dx = –sin(x)
In the following verification of i, we write s(x) for sin(x)
and c(x) for cos(x). We need the following limit–results.
lim
sin(h)
h
= 1
h→0
cos(h) – 1
Expand the difference quotient of sin(x) by the Sum of
Angles Formula for sine.
s(x + h) – s(x)
h =
s(x)c(h) + c(x)s(h) – s(x)
h
= s(x)c(h) – s(x) + c(x)s(h)
h
lim
h
= 0
h→0
(HW)

19. More Computations of Derivatives
Derivatives of Sine and Cosine
i. sin'(x) = cos(x)
d sin(x)
dx = cos(x)
ii. cos'(x) = –sin(x)
d cos(x)
dx = –sin(x)
In the following verification of i, we write s(x) for sin(x)
and c(x) for cos(x). We need the following limit–results.
lim
sin(h)
h
= 1
h→0
cos(h) – 1
Expand the difference quotient of sin(x) by the Sum of
Angles Formula for sine.
s(x + h) – s(x)
h =
s(x)c(h) + c(x)s(h) – s(x)
h
= s(x)c(h) – s(x) + c(x)s(h)
h
= s(x)(c(h) – 1) + c(x)s(h)
h
lim
h
= 0
h→0
(HW)

20. More Computations of Derivatives
Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h

21. More Computations of Derivatives
Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
c(x)s(h)
+ lim
h→0
h

22. More Computations of Derivatives
Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
0 1
c(x)s(h)
+ lim
h→0
h

23. More Computations of Derivatives
Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
0 1
c(x)s(h)
+ lim
h→0
h = cos(x)

24. More Computations of Derivatives
Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
0 1
c(x)s(h)
+ lim
h→0
h = cos(x)
Therefore d sin(x)
dx = cos(x)

25. More Computations of Derivatives
Hence lim
h→0
s(x + h) – s(x)
h
= lim
h→0
s(x)(c(h) – 1) + c(x)s(h)
h
= lim
h→0
s(x)(c(h) – 1)
h
0 1
c(x)s(h)
+ lim
h→0
h = cos(x)
Therefore d sin(x)
dx = cos(x)
The verification of ii is similar and we leave it as an
exercise.

26. More Computations of Derivatives
Derivatives of Other Trig. Functions
The derivatives of the other trig. functions may be
obtained by the quotient rule and the derivatives of
sine and cosine.
tan'(x) = sec2(x)
d tan(x)
dx = sec2(x)
cot'(x) = –csc2(x)
d cot(x)
dx = –csc2(x)
sec'(x) = sec(x)tan(x)
csc'(x) = –csc(x)cot(x)
d sec(x)
dx = sec(x)tan(x)
d csc(x)
dx = –csc(x)cot(x)

27. More Computations of Derivatives
Example A. Find the derivatives of the following
functions.
a.
d tan(x)
dx
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)

28. More Computations of Derivatives
Example A. Find the derivatives of the following
functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)

29. More Computations of Derivatives
Example A. Find the derivatives of the following
functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
=
c(x)s'(x) – s(x)c'(x)
(xN)' = NxN–1
c2(x) sin'(x) = cos(x)
cos'(x) = –sin(x)

30. More Computations of Derivatives
Example A. Find the derivatives of the following
functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
=
c(x)s'(x) – s(x)c'(x)
(xN)' = NxN–1
c2(x) sin'(x) = cos(x)
cos'(x) = –sin(x)
c(x) –s(x)

31. More Computations of Derivatives
Example A. Find the derivatives of the following
functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
=
c(x)s'(x) – s(x)c'(x)
(xN)' = NxN–1
c2(x) sin'(x) = cos(x)
cos'(x) = –sin(x)
c(x) –s(x)
=
c2(x) + s2(x)
c2(x)

32. More Computations of Derivatives
Example A. Find the derivatives of the following
functions.
a.
d tan(x)
dx
d
dx
= s(x)
c(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
=
c(x)s'(x) – s(x)c'(x)
(xN)' = NxN–1
c2(x) sin'(x) = cos(x)
cos'(x) = –sin(x)
c(x) –s(x)
=
c2(x) + s2(x)
c2(x)
=
1
c2(x)
= sec2(x)

33. More Computations of Derivatives
b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)

34. More Computations of Derivatives
b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'

35. More Computations of Derivatives
b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'
s2(x)
=
0 c(x)
s(x)(1)' – 1*s'(x)

36. More Computations of Derivatives
b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'
s2(x)
=
0 c(x)
s(x)(1)' – 1*s'(x)
=
–c(x)
s2(x)
= –csc(x)cot(x)

37. More Computations of Derivatives
b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'
s2(x)
=
0 c(x)
s(x)(1)' – 1*s'(x)
=
–c(x)
s2(x)
= –csc(x)cot(x)
Replacing s(x) by a generic function f(x) in the above
proof we get the a short cut for the derivative of 1/f.

38. More Computations of Derivatives
b. csc'(x)
(fg)' = f 'g + fg'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
1
s(x)
= ( )'
s2(x)
=
0 c(x)
s(x)(1)' – 1*s'(x)
=
–c(x)
s2(x)
= –csc(x)cot(x)
Replacing s(x) by a generic function f(x) in the above
proof we get the a short cut for the derivative of 1/f.
–f '
(Derivative of 1/f) (1/f)' =
f2
d(1/f)
dx
–df/dx
= f2

39. More Computations of Derivatives
Example B. Find the derivatives of (fg)' = f 'g + fg'
(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)

40. More Computations of Derivatives
Example B. Find the derivatives of (fg)' = f 'g + fg'
(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'

41. More Computations of Derivatives
Example B. Find the derivatives of (fg)' = f 'g + fg'
(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= tan'(x)*tan(x) + tan(x)*tan'(x)

42. More Computations of Derivatives
Example B. Find the derivatives of (fg)' = f 'g + fg'
(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= tan'(x)*tan(x) + tan(x)*tan'(x)
= 2tan(x)[tan'(x)]

43. More Computations of Derivatives
Example B. Find the derivatives of (fg)' = f 'g + fg'
(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= tan'(x)*tan(x) + tan(x)*tan'(x)
= 2tan(x)[tan'(x)]
= 2tan(x)sec2(x)

44. More Computations of Derivatives
Example B. Find the derivatives of (fg)' = f 'g + fg'
(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= tan'(x)*tan(x) + tan(x)*tan'(x)
= 2tan(x)[tan'(x)]
= 2tan(x)sec2(x)
Again replacing tan(x) by a function f(x) in the above
proof we get the a short cut for the derivative f2(x).

45. More Computations of Derivatives
Example B. Find the derivatives of (fg)' = f 'g + fg'
(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= tan'(x)*tan(x) + tan(x)*tan'(x)
= 2tan(x)[tan'(x)]
Again replacing tan(x) by a function f(x) in the above
proof we get the a short cut for the derivative f2(x).
(Derivative of f2(x)) (f2(x))' = 2f(x)f '(x)
d(f2)= 2f
dx
df
dx
= 2tan(x)sec2(x)

46. More Computations of Derivatives
Example B. Find the derivatives of (fg)' = f 'g + fg'
(tan2(x))'
gf ' – fg'
g2
f
( g )' =
(xN)' = NxN–1
sin'(x) = cos(x)
cos'(x) = –sin(x)
= (tan(x)*tan(x))'
= tan'(x)*tan(x) + tan(x)*tan'(x)
= 2tan(x)[tan'(x)]
= 2tan(x)sec2(x)
Again replacing tan(x) by a function f(x) in the above
proof we get the a short cut for the derivative f2(x).
(Derivative of f2(x)) (f2(x))' = 2f(x)f '(x)
d(f2)= 2f
dx
df
dx
Note the attached tail end derivative factor f '(x).

47. More Computations of Derivatives
The Chain Rule is the rule for calculating the
derivative of the composition of two functions.

48. More Computations of Derivatives
The Chain Rule is the rule for calculating the
derivative of the composition of two functions.
It demonstrates how the derivative of the composition
can be assembled from the derivatives of the
component functions.

49. More Computations of Derivatives
The Chain Rule is the rule for calculating the
derivative of the composition of two functions.
It demonstrates how the derivative of the composition
can be assembled from the derivatives of the
component functions.
(The Chain Rule) Let u = u(v) and v = v(x),
the derivative of the composition function
(u ○ v)(x) = u( v(x) ) is the product of the derivatives
of u and v,

50. More Computations of Derivatives
The Chain Rule is the rule for calculating the
derivative of the composition of two functions.
It demonstrates how the derivative of the composition
can be assembled from the derivatives of the
component functions.
(The Chain Rule) Let u = u(v) and v = v(x),
the derivative of the composition function
(u ○ v)(x) = u( v(x) ) is the product of the derivatives
of u and v, that is du
dv
dv
dx
du =
dx

51. More Computations of Derivatives
The Chain Rule is the rule for calculating the
derivative of the composition of two functions.
It demonstrates how the derivative of the composition
can be assembled from the derivatives of the
component functions.
(The Chain Rule) Let u = u(v) and v = v(x),
the derivative of the composition function
(u ○ v)(x) = u( v(x) ) is the product of the derivatives
of u and v, that is du
dv
dv
dx
du =
dx
Let’s use the last example again.
Let u = u(v) = v2,
and v = v(x) = tan(x),
the composition is u(v(x)) = tan2(x).

52. More Computations of Derivatives
We have that
u = u(v) = v2, and v = v(x) = tan(x) and
u(v(x)) = tan2(x)

53. More Computations of Derivatives
We have that
u = u(v) = v2, and v = v(x) = tan(x) and
u(v(x)) = tan2(x)
d tan2(x)
dx
=
Hence
du
dv
dv
dx

54. More Computations of Derivatives
We have that
u = u(v) = v2, and v = v(x) = tan(x) and
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
Hence
du
dv
dv
dx

55. More Computations of Derivatives
We have that
u = u(v) = v2, and v = v(x) = tan(x) and
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
d tan(x)
dx
Hence
du
dv
dv
dx

56. More Computations of Derivatives
We have that
u = u(v) = v2, and v = v(x) = tan(x) and
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
d tan(x)
dx
Hence
du
dv
dv
dx
the back–
derivative

57. More Computations of Derivatives
We have that
u = u(v) = v2, and v = v(x) = tan(x) and
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
d tan(x)
dx
Hence
du
dv
dv
dx
the back–
derivative
= (2v) sec2(x)

58. More Computations of Derivatives
We have that
u = u(v) = v2, and v = v(x) = tan(x) and
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
d tan(x)
dx
Hence
du
dv
dv
dx
the back–
derivative
= (2v) sec2(x)
= 2tan(x) sec2(x)

59. More Computations of Derivatives
We have that
u = u(v) = v2, and v = v(x) = tan(x) and
u(v(x)) = tan2(x)
d tan2(x)
dx
= dv2
dv
=
d tan(x)
dx
Hence
du
dv
dv
dx
= (2v) sec2(x)
the back–
derivative
= 2tan(x) sec2(x)
The name “Chain Rule” came from the fact that the
rule is often applied many times with the
back–derivative spinning off another back–derivative
which spins off another etc…

60. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
b. Let y = sin3(x) find dy/dx.

61. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
We have that y = sin(u) = sin(x3).
b. Let y = sin3(x) find dy/dx.

62. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
We have that y = sin(u) = sin(x3).
dy
dx =
dy
du
du
dx
Hence
b. Let y = sin3(x) find dy/dx.

63. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
We have that y = sin(u) = sin(x3).
dy
dx =
dy
du
du
dx =
dsin(u)
du
Hence
b. Let y = sin3(x) find dy/dx.

64. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
We have that y = sin(u) = sin(x3).
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
Hence
b. Let y = sin3(x) find dy/dx.
the back–
derivative

65. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
We have that y = sin(u) = sin(x3).
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
Hence
b. Let y = sin3(x) find dy/dx.
the back–
derivative

66. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
We have that y = sin(u) = sin(x3).
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
Hence
b. Let y = sin3(x) find dy/dx.
We have that y = u3 where u = sin(x).
the back–
derivative

67. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
We have that y = sin(u) = sin(x3).
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
b. Let y = sin3(x) find dy/dx.
We have that y = u3 where u = sin(x).
dy
dx =
dy
du
du
dx
Hence
Hence
the back–
derivative

68. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
We have that y = sin(u) = sin(x3).
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
b. Let y = sin3(x) find dy/dx.
We have that y = u3 where u = sin(x).
dy
dx =
dy
du
du
dx =
du3
du
d sin(x)
dx
Hence
Hence
the back–
derivative
the back–
derivative

69. More Computations of Derivatives
Example C.
a. Let y = sin(u) and u = x3, express y as a function
in x and find dy/dx.
We have that y = sin(u) = sin(x3).
dy
dx =
dy
du
du
dx =
dsin(u)
du
dx3
dx
= cos(u) (3x2) = 3x2cos(x3)
b. Let y = sin3(x) find dy/dx.
We have that y = u3 where u = sin(x).
dy
dx =
dy
du
du
dx =
du3
du
d sin(x)
dx
= 3u2 cos(x) = 3sin2(x) cos(x)
Hence
Hence
the back–
derivative
the back–
derivative

70. More Computations of Derivatives
(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dy
du
=
..
dv
dx du
dw
dx

71. More Computations of Derivatives
(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dy
du
=
..
dv
dx du
dw
dx
c. Find the derivative of cos[(x2+1)3]

72. More Computations of Derivatives
(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dy
du
=
..
dv
dx du
dw
dx
c. Find the derivative of cos[(x2+1)3]
We have that y = cos(u),

73. More Computations of Derivatives
(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dy
du
=
..
dv
dx du
dw
dx
c. Find the derivative of cos[(x2+1)3]
We have that y = cos(u),
where u = v3,

74. More Computations of Derivatives
(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dy
du
=
..
dv
dx du
dw
dx
c. Find the derivative of cos[(x2+1)3]
We have that y = cos(u),
where u = v3,
where v = (x2 + 1).

75. More Computations of Derivatives
(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dy
du
=
..
dv
dx du
dw
dx
c. Find the derivative of cos[(x2+1)3]
We have that y = cos(u),
where u = v3,
where v = (x2 + 1).
dy
dx =
dy
du
du
dv
dv
Hence dx

76. More Computations of Derivatives
(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dx =
c. Find the derivative of cos[(x2+1)3]
We have that y = cos(u),
where u = v3,
where v = (x2 + 1).
dy
dx =
dy
du
du
dv
dv
Hence dx
= –sin(u) (3v2) (2x)
dy
du
du
dw
dv
dx
..

77. More Computations of Derivatives
(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dx =
c. Find the derivative of cos[(x2+1)3]
We have that y = cos(u),
where u = v3,
where v = (x2 + 1).
dy
dx =
dy
du
du
dv
dv
Hence dx
= –sin(u) (3v2) (2x)
= –sin(v3) (3v2) (2x)
dy
du
du
dw
dv
dx
..

78. More Computations of Derivatives
(The General Chain Rule)
If y = y(u), u = u(w), .. v = v(x) then
dy
dx =
c. Find the derivative of cos[(x2+1)3]
We have that y = cos(u),
where u = v3,
where v = (x2 + 1).
dy
dx =
dy
du
du
dv
dv
Hence dx
= –sin(u) (3v2) (2x)
= –sin(v3) (3v2) (2x)
= –sin[(x2 + 1)3] [3(x2 + 1)2](2x)
dy
du
du
dw
dv
dx
..

79. More Computations of Derivatives
The prime notation of the Chain Rule may look
confusing.
(u ○ v)'(x) = u'(v(x))v'(x)

80. More Computations of Derivatives
The prime notation of the Chain Rule may look
confusing.
derivative taken
with respect to x
derivative taken
with respect to x
derivative taken
with respect to v
(u ○ v)'(x) = u'(v(x))v'(x)

81. More Computations of Derivatives
The prime notation of the Chain Rule may look
confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
is the derivative of u with
dv v=v(x)
respect to v, evaluated at v (x).
derivative taken
with respect to x
derivative taken
with respect to v

82. More Computations of Derivatives
The prime notation of the Chain Rule may look
confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
is the derivative of u with
dv v=v(x)
respect to v, evaluated at v (x).
derivative taken
with respect to x
derivative taken
with respect to v
Example D. Let u(v) = cos(v) and v(x) = x2, find the
slope of the tangent at x = √π for (u ○ v)(x).

83. More Computations of Derivatives
The prime notation of the Chain Rule may look
confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
is the derivative of u with
dv v=v(x)
respect to v, evaluated at v (x).
derivative taken
with respect to x
derivative taken
with respect to v
Example D. Let u(v) = cos(v) and v(x) = x2, find the
slope of the tangent at x = √π for (u ○ v)(x).
(u ○ v)'(x) = u'(v(x))v'(x) =

84. More Computations of Derivatives
The prime notation of the Chain Rule may look
confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
is the derivative of u with
dv v=v(x)
respect to v, evaluated at v (x).
derivative taken
with respect to x
derivative taken
with respect to v
Example D. Let u(v) = cos(v) and v(x) = x2, find the
slope of the tangent at x = √π for (u ○ v)(x).
(u ○ v)'(x) = u'(v(x))v'(x) = –sin(v) 2x,

85. More Computations of Derivatives
The prime notation of the Chain Rule may look
confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
is the derivative of u with
dv v=v(x)
respect to v, evaluated at v (x).
derivative taken
with respect to x
derivative taken
with respect to v
Example D. Let u(v) = cos(v) and v(x) = x2, find the
slope of the tangent at x = √π for (u ○ v)(x).
(u ○ v)'(x) = u'(v(x))v'(x) = –sin(v) 2x, at x = √π,
(u ○ v)'(√π) = –sin(v) 2x x = √π,

86. More Computations of Derivatives
The prime notation of the Chain Rule may look
confusing.
derivative taken
with respect to x
(u ○ v)'(x) = u'(v(x))v'(x)
where u'(v(x)) = du
is the derivative of u with
dv v=v(x)
respect to v, evaluated at v (x).
derivative taken
with respect to x
derivative taken
with respect to v
Example D. Let u(v) = cos(v) and v(x) = x2, find the
slope of the tangent at x = √π for (u ○ v)(x).
(u ○ v)'(x) = u'(v(x))v'(x) = –sin(v) 2x, at x = √π,
(u ○ v)'(√π) = –sin(v) 2x x = √π,
= –sin(√π2) 2√π = 0

87. More Computations of Derivatives
The Chain Rule gives the multiplying effect of the
rates of change under the composition operation.

88. More Computations of Derivatives
The Chain Rule gives the multiplying effect of the
rates of change under the composition operation.
Suppose y = g(u) = 3u and u = f(x) = 5x.

89. More Computations of Derivatives
The Chain Rule gives the multiplying effect of the
rates of change under the composition operation.
Suppose y = g(u) = 3u and u = f(x) = 5x.
The slope of y = g(u) = 3u as a function in u is 3.
The slope u = 5x with respect to x is 5.

90. More Computations of Derivatives
The Chain Rule gives the multiplying effect of the
rates of change under the composition operation.
Suppose y = g(u) = 3u and u = f(x) = 5x.
The slope of y = g(u) = 3u as a function in u is 3.
The slope u = 5x with respect to x is 5.
The composition y = g(f(x)) as a function in x is
y = 3(5x) = 15x which has slope to x as (3)(5) = 15.