1. Chain Rule Short Cuts

2. Chain Rule Short Cuts
Set u = u(x) and y = up where p is a number, or
y = cos(u), or y = sin(u), we get the following rules.

3. Chain Rule Short Cuts
Set u = u(x) and y = up where p is a number, or
y = cos(u), or y = sin(u), we get the following rules.
(Specific Chain Rules)
a. [up]' = pup–1(u)'
b. [sin(u)]' = cos(u)*(u)'
c. [cos(u)]' = –sin(u)*(u)'
(The Power Chain Rule)
(The Sine Chain Rule)
(The Cosine Chain Rule)

4. Chain Rule Short Cuts
Set u = u(x) and y = up where p is a number, or
y = cos(u), or y = sin(u), we get the following rules.
(Specific Chain Rules)
a. [up]' = pup–1(u)'
(The Power Chain Rule)
b. [sin(u)]' = cos(u)*(u)'
(The Sine Chain Rule)
c. [cos(u)]' = –sin(u)*(u)'
(The Cosine Chain Rule)
where ( u )' is the back-derivative with respect to x.

5. Chain Rule Short Cuts
Set u = u(x) and y = up where p is a number, or
y = cos(u), or y = sin(u), we get the following rules.
(Specific Chain Rules)
a. [up]' = pup–1(u)'
(The Power Chain Rule)
b. [sin(u)]' = cos(u)*(u)'
(The Sine Chain Rule)
c. [cos(u)]' = –sin(u)*(u)'
(The Cosine Chain Rule)
where ( u )' is the back-derivative with respect to x.
Example A. a. Find the derivative of (x3+1)2.

6. Chain Rule Short Cuts
Set u = u(x) and y = up where p is a number, or
y = cos(u), or y = sin(u), we get the following rules.
(Specific Chain Rules)
a. [up]' = pup–1(u)'
(The Power Chain Rule)
b. [sin(u)]' = cos(u)*(u)'
(The Sine Chain Rule)
c. [cos(u)]' = –sin(u)*(u)'
(The Cosine Chain Rule)
where ( u )' is the back-derivative with respect to x.
Example A. a. Find the derivative of (x3+1)2.
Set the base u = (x3+1), p = 2 and use the
Power Rule [up]' = pup–1(u)'

7. Chain Rule Short Cuts
Set u = u(x) and y = up where p is a number, or
y = cos(u), or y = sin(u), we get the following rules.
(Specific Chain Rules)
a. [up]' = pup–1(u)'
(The Power Chain Rule)
b. [sin(u)]' = cos(u)*(u)'
(The Sine Chain Rule)
c. [cos(u)]' = –sin(u)*(u)'
(The Cosine Chain Rule)
where ( u )' is the back-derivative with respect to x.
Example A. a. Find the derivative of (x3+1)2.
Set the base u = (x3+1), p = 2 and use the
Power Rule [up]' = pup–1(u)' to get
[(x3+1)2]'

8. Chain Rule Short Cuts
Set u = u(x) and y = up where p is a number, or
y = cos(u), or y = sin(u), we get the following rules.
(Specific Chain Rules)
a. [up]' = pup–1(u)'
(The Power Chain Rule)
b. [sin(u)]' = cos(u)*(u)'
(The Sine Chain Rule)
c. [cos(u)]' = –sin(u)*(u)'
(The Cosine Chain Rule)
where ( u )' is the back-derivative with respect to x.
Example A. a. Find the derivative of (x3+1)2.
Set the base u = (x3+1), p = 2 and use the
Power Rule [up]' = pup–1(u)' to get
[(x3+1)2]'
= 2(x3+1)2–1

9. Chain Rule Short Cuts
Set u = u(x) and y = up where p is a number, or
y = cos(u), or y = sin(u), we get the following rules.
(Specific Chain Rules)
a. [up]' = pup–1(u)'
(The Power Chain Rule)
b. [sin(u)]' = cos(u)*(u)'
(The Sine Chain Rule)
c. [cos(u)]' = –sin(u)*(u)'
(The Cosine Chain Rule)
where ( u )' is the back-derivative with respect to x.
Example A. a. Find the derivative of (x3+1)2.
Set the base u = (x3+1), p = 2 and use the
Power Rule [up]' = pup–1(u)' to get
[(x3+1)2]'
= 2(x3+1)2–1(x3+1)'
u'

10. Set u = u(x) and y = up where p is a number, or
y = cos(u), or y = sin(u), we get the following rules.
(Specific Chain Rules)
a. [up]' = pup–1(u)'
b. [sin(u)]' = cos(u)*(u)'
c. [cos(u)]' = –sin(u)*(u)'
where ( u )' is the back-derivative with respect to x.
Example A. a. Find the derivative of (x3+1)2.
Set the base u = (x3+1), p = 2 and use the
Power Rule [up]' = pup–1(u)' to get
= 6x2 (x3+1)
Chain Rule Short Cuts
= 2(x3+1)3x2
(The Power Chain Rule)
(The Sine Chain Rule)
(The Cosine Chain Rule)
[(x3+1)2]'
= 2(x3+1)2–1(x3+1)'
u'

11. Chain Rule Short Cuts
b. Find the derivative of cos[(x3+1)2]

12. Chain Rule Short Cuts
b. Find the derivative of cos[(x3+1)2]
Set the cosine input (x3+1)2 = u, and use the
Cosine Rule cos(u)' = –sin(u)(u)'

13. Chain Rule Short Cuts
b. Find the derivative of cos[(x3+1)2]
Set the cosine input (x3+1)2 = u, and use the
Cosine Rule cos(u)' = –sin(u)(u)' to get
(cos[(x3+1)2])' =

14. Chain Rule Short Cuts
b. Find the derivative of cos[(x3+1)2]
Set the cosine input (x3+1)2 = u, and use the
Cosine Rule cos(u)' = –sin(u)(u)' to get
(cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]'

15. Chain Rule Short Cuts
b. Find the derivative of cos[(x3+1)2]
Set the cosine input (x3+1)2 = u, and use the
Cosine Rule cos(u)' = –sin(u)(u)' to get
(cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]'
the new “u” for u2

16. Chain Rule Short Cuts
b. Find the derivative of cos[(x3+1)2]
Set the cosine input (x3+1)2 = u, and use the
Cosine Rule cos(u)' = –sin(u)(u)' to get
(cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]'
the new “u” for u2
= –sin[(x3+1)2] [2(x3+1)1](x3+1)'

17. Chain Rule Short Cuts
b. Find the derivative of cos[(x3+1)2]
Set the cosine input (x3+1)2 = u, and use the
Cosine Rule cos(u)' = –sin(u)(u)' to get
(cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]'
the new “u” for u2
= –sin[(x3+1)2] [2(x3+1)1](x3+1)'
= –sin[(x3+1)2] [2(x3+1)]3x2

18. Chain Rule Short Cuts
b. Find the derivative of cos[(x3+1)2]
Set the cosine input (x3+1)2 = u, and use the
Cosine Rule cos(u)' = –sin(u)(u)' to get
(cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]'
the new “u” for u2
= –sin[(x3+1)2] [2(x3+1)1](x3+1)'
= –sin[(x3+1)2] [2(x3+1)]3x2
We list the frequently used Chain Rules below.

19. Chain Rule Short Cuts
b. Find the derivative of cos[(x3+1)2]
Set the cosine input (x3+1)2 = u, and use the
Cosine Rule cos(u)' = –sin(u)(u)' to get
(cos[(x3+1)2])' = –sin[(x3+1)2] [(x3+1)2]'
the new “u” for u2
= –sin[(x3+1)2] [2(x3+1)1](x3+1)'
= –sin[(x3+1)2] [2(x3+1)]3x2
We list the frequently used Chain Rules below.
The list includes the log and exponential functions
for the sole purpose of practicing symbolic algebraic
manipulation. The verifications will be given later.

20. Chain Rule Short Cuts
(Chain Rules Short Cuts)
(The Power Chain Rule)
[up]' = pup–1(u)'
(The Trig. Chain Rules)
[sin(u)] ' = sin'(u) = cos(u)(u)'
[cos(u)] ' = cos'(u) = –sin(u)(u)'
[tan(u)] ' = tan'(u) = sec2(u)(u)'
[cot(u)] ' = cot'(u) = –csc2(u)(u)'
[sec (u)] ' = sec'(u) = sec(u)tan(u)(u)'
[csc (u)] ' = csc'(u) = –csc(u)cot(u)(u)'
(The Derivatives of Log and Exponential Functions)
and [eu]' = eu(u)'
and [ln(u)]' = (u)'
u
[ex]' = ex
[ln(x)]' = x 1

21. dup
dx
d sin(u)
dx
Chain Rule Short Cuts
Below are the d/dx versions of the same rules.
= pup–1du
dx
= cos(u) du
dx
d cos(u)
dx = –sin(u) du
dx
d tan(u)
dx
= sec2(u)du
dx
d cot(u)
dx = –csc2(u) du
dx
d sec(u)
dx
= sec(u)tan(u)du
dx
d csc(u)
dx = –csc(u)cot(u)du
dx
deu
dx = eu du
dx
d ln(u)
dx =
du
1
u dx

22. Chain Rule Short Cuts
Usually repeated applications of the above rules are
required to compute the derivative of an arbitrary
elementary function.

23. Chain Rule Short Cuts
Usually repeated applications of the above rules are
required to compute the derivative of an arbitrary
elementary function. The particular chain rule to be
used at a particular step corresponds to the last
function applied in the expression.

24. Chain Rule Short Cuts
Usually repeated applications of the above rules are
required to compute the derivative of an arbitrary
elementary function. The particular chain rule to be
used at a particular step corresponds to the last
function applied in the expression.
Example B. Find the following derivatives.
a. y = sin(x3)

25. Chain Rule Short Cuts
Usually repeated applications of the above rules are
required to compute the derivative of an arbitrary
elementary function. The particular chain rule to be
used at a particular step corresponds to the last
function applied in the expression.
Example B. Find the following derivatives.
a. y = sin(x3)
The formula takes the input x→x3→sin(x3).

26. Chain Rule Short Cuts
Usually repeated applications of the above rules are
required to compute the derivative of an arbitrary
elementary function. The particular chain rule to be
used at a particular step corresponds to the last
function applied in the expression.
Example B. Find the following derivatives.
a. y = sin(x3)
The formula takes the input x→x3→sin(x3).
The last operation applied is sin(u) so we view
sin(x3) as sin(u) with u = x3.

27. Chain Rule Short Cuts
Usually repeated applications of the above rules are
required to compute the derivative of an arbitrary
elementary function. The particular chain rule to be
used at a particular step corresponds to the last
function applied in the expression.
Example B. Find the following derivatives.
a. y = sin(x3)
The formula takes the input x→x3→sin(x3).
The last operation applied is sin(u) so we view
sin(x3) as sin(u) with u = x3.
Hence dx = cos(x3) d sin(x3)

28. Chain Rule Short Cuts
Usually repeated applications of the above rules are
required to compute the derivative of an arbitrary
elementary function. The particular chain rule to be
used at a particular step corresponds to the last
function applied in the expression.
Example B. Find the following derivatives.
a. y = sin(x3)
The formula takes the input x→x3→sin(x3).
The last operation applied is sin(u) so we view
sin(x3) as sin(u) with u = x3.
dx = d x3
dx
Hence d sin(x cos(x3) 3)

29. Chain Rule Short Cuts
Usually repeated applications of the above rules are
required to compute the derivative of an arbitrary
elementary function. The particular chain rule to be
used at a particular step corresponds to the last
function applied in the expression.
Example B. Find the following derivatives.
a. y = sin(x3)
The formula takes the input x→x3→sin(x3).
The last operation applied is sin(u) so we view
sin(x3) as sin(u) with u = x3.
dx = d x3
dx
Hence d sin(x cos(x3) 3)
= 3x2cos(x3)

30. Chain Rule Short Cuts
b. y = sin3(x)

31. Chain Rule Short Cuts
b. y = sin3(x)
The formula takes the input x→sin(x)→[sin(x)]3
so the last operation applied is u3 where u = sin(x).

32. Chain Rule Short Cuts
b. y = sin3(x)
The formula takes the input x→sin(x)→[sin(x)]3
so the last operation applied is u3 where u = sin(x).
Using the Power Rule with u3 we have
[sin3(x)]'
= [u3]'

33. Chain Rule Short Cuts
b. y = sin3(x)
The formula takes the input x→sin(x)→[sin(x)]3
so the last operation applied is u3 where u = sin(x).
Using the Power Rule with u3 we have
[sin3(x)]'
= [u3]'
= 3u2[u]'

34. Chain Rule Short Cuts
b. y = sin3(x)
The formula takes the input x→sin(x)→[sin(x)]3
so the last operation applied is u3 where u = sin(x).
Using the Power Rule with u3 we have
[sin3(x)]'
= [u3]'
= 3u2[u]'
= 3sin2(x)[sin(x)]'

35. Chain Rule Short Cuts
b. y = sin3(x)
The formula takes the input x→sin(x)→[sin(x)]3
so the last operation applied is u3 where u = sin(x).
Using the Power Rule with u3 we have
[sin3(x)]'
= [u3]'
= 3u2[u]'
= 3sin2(x)[sin(x)]'
= 3sin2(x)cos(x)

36. Chain Rule Short Cuts
b. y = sin3(x)
The formula takes the input x→sin(x)→[sin(x)]3
so the last operation applied is u3 where u = sin(x).
Using the Power Rule with u3 we have
[sin3(x)]'
= [u3]'
= 3u2[u]'
= 3sin2(x)[sin(x)]'
= 3sin2(x)cos(x)
c. y = ln(5e3x – 2)

37. Chain Rule Short Cuts
b. y = sin3(x)
The formula takes the input x→sin(x)→[sin(x)]3
so the last operation applied is u3 where u = sin(x).
Using the Power Rule with u3 we have
[sin3(x)]'
= [u3]'
= 3u2[u]'
= 3sin2(x)[sin(x)]'
= 3sin2(x)cos(x)
c. y = ln(5e3x – 2)
The last operation applied is the log–function
so we use the Log Chain Rule with u = (5e3x – 2).

38. Chain Rule Short Cuts
d ln(5e3x – 2)
dx
d ln(u)
dx
=
du
1
u dx
deu
dx
= eu du
dx

39. Chain Rule Short Cuts
d ln(5e3x – 2)
dx
=
1 d (5e3x – 2)
(5e3x – 2)
dx
d ln(u)
dx
=
du
1
u dx
deu
dx
= eu du
dx

40. Chain Rule Short Cuts
d ln(5e3x – 2)
dx
=
1 d (5e3x – 2)
(5e3x – 2)
dx
=
1 5 d (e3x)
(5e3x – 2)
dx
d ln(u)
dx
=
du
1
u dx
deu
dx
= eu du
dx

41. Chain Rule Short Cuts
d ln(5e3x – 2)
dx
=
1 d (5e3x – 2)
(5e3x – 2)
dx
=
1 5 d (e3x)
(5e3x – 2)
dx
d ln(u)
dx
=
du
1
u dx
the new u
deu
dx
= eu du
dx

42. Chain Rule Short Cuts
d ln(5e3x – 2)
dx
=
1 d (5e3x – 2)
(5e3x – 2)
dx
=
1 5 d (e3x)
(5e3x – 2)
dx
d ln(u)
dx
=
du
1
u dx
the new u
deu
dx
= eu du
dx
=
(5e3x – 2)
d 3x
dx
5 e3x

43. Chain Rule Short Cuts
d ln(5e3x – 2)
dx
=
1 d (5e3x – 2)
(5e3x – 2)
dx
=
1 5 d (e3x)
(5e3x – 2)
dx
d ln(u)
dx
=
du
1
u dx
the new u
deu
dx
= eu du
dx
=
(5e3x – 2)
d 3x
dx
5 e3x
=
15 e3x
(5e3x – 2)

44. Chain Rule Short Cuts
d ln(5e3x – 2)
dx
=
1 d (5e3x – 2)
(5e3x – 2)
dx
=
1 5 d (e3x)
(5e3x – 2)
dx
d ln(u)
dx
=
du
1
u dx
the new u
deu
dx
= eu du
dx
=
5 e3x
(5e3x – 2)
d 3x
dx
=
15 e3x
(5e3x – 2)
If the last operation applied in the expression were the
algebraic operations of multiplication or division by a
non–constant, then the Product and Quotient Rules
should be applied respectively.

45. Chain Rule Short Cuts
d. y =
ex2+1
ln(x)

46. Chain Rule Short Cuts
d. y =
ex2+1
ln(x)
The last operation executed in the formula is division
so we use the Quotient Rule first.

47. Chain Rule Short Cuts
d. y =
ex2+1
ln(x)
The last operation executed in the formula is division
so we use the Quotient Rule first.
ex2+1
ln(x)
[ ]' gf ' – fg'
g2
f
g
( )' =

48. Chain Rule Short Cuts
d. y =
ex2+1
ln(x)
The last operation executed in the formula is division
so we use the Quotient Rule first.
ex2+1
ln(x)
[ ]'
=
x2+1 ln(x)[e ]' –
ln2(x)
gf ' – fg'
g2
f
g
( )' =
ex2+1[ln(x)]'

49. Chain Rule Short Cuts
d. y =
ex2+1
ln(x)
The last operation executed in the formula is division
so we use the Quotient Rule first.
ex2+1
ln(x)
[ ]'
=
x2+1 ln(x)[e ]' –
ex2+1[ln(x)]'
ln2(x)
1/x
gf ' – fg'
g2
f
g
( )' =

50. Chain Rule Short Cuts
d. y =
ex2+1
ln(x)
The last operation executed in the formula is division
so we use the Quotient Rule first.
ex2+1
ln(x)
[ ]'
=
Use the
Exponential
Chain Rule
x2+1 ln(x)[e ]' –
ex2+1[ln(x)]'
ln2(x)
1/x
gf ' – fg'
g2
f
g
( )' =

51. Chain Rule Short Cuts
d. y =
ex2+1
ln(x)
The last operation executed in the formula is division
so we use the Quotient Rule first.
ex2+1
ln(x)
[ ]'
=
x2+1 ln(x)[e ]' –
ex2+1[ln(x)]'
ln2(x)
=
x2+1 ln(x)[e ][x2+1]' –
ln2(x)
ex2+1/x
Use the
Exponential
Chain Rule
1/x
u'
=
gf ' – fg'
g2
f
g
( )' =

52. Chain Rule Short Cuts
d. y =
ex2+1
ln(x)
The last operation executed in the formula is division
so we use the Quotient Rule first.
ex2+1
ln(x)
[ ]'
=
x2+1 ln(x)[e ]' –
ex2+1[ln(x)]'
ln2(x)
=
x2+1 ln(x)[e ][x2+1]' –
ln2(x)
ex2+1/x
Use the
Exponential
Chain Rule
1/x
u'
=
gf ' – fg'
x2+1
e [2x2ln(x) – 1]
xln2(x)
g2
f
g
( )' =