CHAPTER 3. RULES FOR DERIVATIVES
3.4
76
The Chain rule
Rule (Chain Rule.).
d
f (g(x)) = f � (g(x)) · g � (x)
dx
dy
dy du
=
·
dx
du dx
where y is a function of u and u is a function of x
d
du
f (u) = f � (u) ·
dx
dx
d
f (g(x)) = the derivative of the outside function (don’t
dx
change the inside) times the derivative of the
inside function.
Rule (Chain Rule function by function).
ordinary rule
−→
chain rule version in boxes
d n
x = nxn−1
dx
−→
d
(
dx
d x
e = ex
dx
−→
d
e
dx
d
sin(x) = cos(x)
dx
−→
d
sin(
dx
d
cos(x) = − sin(x)
dx
−→
d
cos(
dx
d
tan(x) = sec2 (x)
dx
−→
d
tan(
dx
�)
n
= n(
�)
n−1
� = e � · ( �)
�
·(
�)
�
�) = cos(�) · (�)
�
�) = − sin(�) · (�)
�) = sec (�) · (�)
2
�
�
or
chain rule version in u
or
d n
du
(u ) = nun−1 ·
dx
dx
or
u du
d u
e =e ·
dx
dx
or
d
du
sin(u) = cos(u) ·
dx
dx
or
d
du
cos(u) = − sin(u) ·
dx
dx
or
d
du
tan(u) = sec2 (u) ·
dx
dx
In each formula, we imagine putting something (some function of x) inside of each
or each u.
�
d
Example 1. Find
sin(x2 + 1) using three or four different kinds of notation,
dx
and decide which kind of notation you like best.
Solution. Eff-prime notation:
sin(x2 + 1) = f (g(x)) where f (x) = sin(x), g(x) = x2 + 1
f � (x) = cos(x)
g � (x) = 2x
f � (g(x)) · g � (x) = cos(x2 + 1) · 2x.
CHAPTER 3. RULES FOR DERIVATIVES
77
Leibniz notation:
sin(x2 + 1) = y(u) where y(u) = sin(u), u = x2 + 1
dy
= cos(u)
du
du
= 2x
dx
dy
dy du
=
= cos(u)2x = cos(x2 + 1)2x.
dx
du dx
Word notation:
d
sin(x2 + 1) = cos(x2 + 1) · 2x
dx
deriv. of
outside
deriv. of
inside
don’t
change
inside
.
Box notation:
�
� �
d
�
sin( ) = cos( ) ·
dx
�
�
�
�
�
d
sin x2 + 1 = cos x2 + 1 · x2 + 1
dx
= cos(x2 + 1) · 2x.
U notation:
d
du
sin(u) = cos(u) ·
dx
dx
d
du
sin(x2 + 1) = cos(u) ·
(where u = x2 + 1)
dx
dx
= cos(x2 + 1) · 2x.
d� 2
4x + x.
dx
Solution. This time we use only three kinds of notation:
U notation:
d√
1
d√
1
du
x = √ −→
u= √ ·
dx
dx
2 x
2 u dx
d� 2
1
du
4x + x = √ ·
dx
2 u dx
1
= 2
· (8x + 1)
4x + x
Example 2. Find
Box notation:
d√
1
d�
x = √ −→
dx
dx
2 x
� = 2�1� · �
�
CHAPTER 3. RULES FOR DERIVATIVES
d
dx
�
78
�
1
4x2 + x = �
· 4x2 + x
2 4x2 + 1
1
= √
· (8x + 1)
2 4x2 + 1
Word notation:
d � 2
1
4x + x = √
· (8x + 1)
dx
2 4x2 + x
deriv. of
outside
d
Example 3. Find
dt
�
deriv. of
inside
don’t
change
inside
sin(t)
2
t +t+1
�15
Solution. We combine the chain rule form of the derivative of (
quotient rule on the inside to get
d
dt
�
d
(
dt
�)
sin(t)
2
t +t+1
15
�15
= 15(
�) · �
14
�
�)
15
and use the
�
�
�14
sin(t)
sin(t)
= 15
· 2
t2 + t + 1
t +t+1
�
�14
sin(t)
cos(t)(t2 + t + 1) − sin(t)(2t + 1)
= 15 2
·
t +t+1
(t2 + t + 1)2
Here’s another way to write it:
d
dt
�
sin(t)
t2 + t + 1
�15
deriv. of
outside
Example 4. Find
= 15
�
sin(t)
t2 + t + 1
�14
·
cos(t)(t2 + t + 1) − sin(t)(2t + 1)
(t2 + t + 1)2
don’t
change
inside
deriv. of
inside
d 2
5x sin(6x2 ).
dx
Solution. We combine the product rule with the chain rule:
�
��
d 2
5x sin(6x2 ) = (5x2 )� sin(6x2 ) + (5x2 ) sin(6x2 )
dx
= 10x sin(6x2 ) + 5x2 cos(6x2 )(6x2 )�
= 10x sin(6x2 ) + 5x2 cos(6x2 )12x
= 10x sin(6x2 ) + 60x3 cos(6x2 )
CHAPTER 3. RULES FOR DERIVATIVES
Example 5. Find e
79
�√
�
5x+3 sin
2 tan(x)+1
Solution.
�√
d 5x+3 sin
e
dx
=e
2 tan(x)+1
5x + 3 sin
�
��
�
2 tan(x) + 1
=e
�√
�
5x+3 sin
2 tan(x)+1
=e
�√
�
5x+3 sin
2 tan(x)+1
=e
�√
�
5x+3 sin
2 tan(x)+1
·
�
��
�
· 5x + 3 sin
2 tan(x) + 1
5 + 3 cos

· 5 + 3 cos
·
�
�
�
2 tan(x) + 1
��
2 tan(x) + 1
�
�
2
�
�
2 tan(x) + 1
�
1
2 tan(x) + 1
��
Rule.
General exponential rule:
d x
a = ax ln(a), for a > 0
dx
Proof.
a = eln(a)
d x
d ln(a) x
a =
(e
)
dx
dx
d x ln(a)
=
e
dx
=e
x ln(a)
· x ln(a)
= ex ln(a) · ln(a)
= (eln(a) )x · ln(a)
= ax ln(a)
�

· 2 tan(x) + 1 
��
�
1
5 + 3 cos
2 tan(x) + 1 �
· 2 sec2 (x)
2 2 tan(x) + 1
This is where we ended on Tuesday, October 15
�
�