Section 2.5 The Chain Rule
2010 Kiryl Tsishchanka
The Chain Rule
PROBLEM: Let f (x) = (1 + x)2 . Find f ′ (x).
Solution 1: To find the derivative of this function, we do algebra first and then apply calculus
rules:
f ′ (x) = [(1 + x)2 ]′ = (1 + 2x + x2 )′ = 1′ + 2(x)′ + (x2 )′ = 0 + 2 · 1 + 2x = 2 + 2x
Solution 2(?): One can try to use the power rule immediately:
f ′ (x) = [(1 + x)2 ]′ = 2(1 + x)2−1 = 2(1 + x)
Note that in both cases we got the same result. However, the goal of Section 2.5 is to show
that despite the fact that Solution 2 gives the right answer, it is not completely correct. To
explain what me mean by that, let us consider the following example:
PROBLEM: Let f (x) = (1 − x)2 . Find f ′ (x).
8
Solution 1: We have
6
2 ′
2 ′
2 ′
f (x) = [(1 − x) ] = (1 − 2x + x ) = 1 − 2(x) + (x )
′
′
′
y
4
= 0 − 2 · 1 + 2x = −2 + 2x
2
Solution 2(???): If we apply the power rule immediately,
we get
0
-2
-1
0
1
2
3
4
x
?
f ′ (x) = [(1 − x)2 ]′ = 2(1 − x)2−1 = 2(1 − x)
-2
Note that we got two different answers. One can easily see that the second answer is incorrect.
In fact, if f ′ (x) = 2(1 − x), then f ′ (2) = 2(1 − 2) = 2(−1) = −2. This means that the slope of
the tangent line to the curve f (x) = (1 − x)2 at x = 2 is negative. But this is not the case!
CONCLUSION: We can’t always apply the rules (xn )′ = nxn−1 , (sin x)′ = cos x, etc. to cases
when we have u instead of x, where u is an algebraic expression different from x.
EXAMPLES:
(x3 )′ = 3x2 ,
[(2 + x)3 ]′ = 3(2 + x)2
(sin x)′ = cos x, [sin(x − 5)]′ = cos(x − 5)
√
1
( x)′ = √ ,
2 x
√
1
( x − 3)′ = √
2 x−3
[(2x)3 ]′ 6= 3(2x)2 ,
[(2 − x)3 ]′ 6= 3(2 − x)2
[sin(4x)]′ 6= cos(4x), [sin(5 − x)]′ 6= cos(5 − x)
√
1
( 5x)′ 6= √ ,
2 5x
1
√
1
( 3 − x)′ 6= √
2 3−x
Section 2.5 The Chain Rule
2010 Kiryl Tsishchanka
THE CHAIN RULE: If f and g are both differentiable and F = f ◦ g is the composite function
defined by F (x) = f (g(x)), then F is differentiable and F ′ is given by the product
F ′ (x) = f ′ (g(x)) · g ′ (x)
In Leibniz notation, if y = f (u) and u = g(x) are both differentiable functions, then
dy
dy du
=
dx
du dx
EXAMPLE: If F (x) = (1 − x)2 , then
F ′ (x) = [(1 − x)2 ]′ = [F = f ◦ g where f (x) = x2 , g(x) = 1 − x] = 2(1 − x) · (1 − x)′
= 2(1 − x)(−1)
= −2(1 − x)
or
d(u2 ) d(1 − x)
d((1 − x)2 )
= [y = u2 , u = 1 − x] =
= 2u · (−1) = 2(1 − x)(−1) = −2(1 − x)
dx
du
dx
DIFFERENTIATION RULES
c′ = 0
(un )′ = nun−1 · u′
[cf (x)]′ = cf ′ (x)
(sin u)′ = cos u · u′
(csc u)′ = − csc u cot u · u′
[f (x) ± g(x)]′ = f ′ (x) ± g ′ (x)
(cos u)′ = − sin u · u′ (sec u)′ = sec u tan u · u′
2
(tan u) = sec u · u
′
′
2
(cot u) = − csc u · u
′
′
[f (x)g(x)]′ = f ′ (x)g(x) + f (x)g ′ (x)
f (x)
g(x)
′
=
f ′ (x)g(x) − f (x)g ′ (x)
[g(x)]2
EXAMPLES:
1. [sin x]′ = cos x · x′ = cos x · 1 = cos x
2. [sin(x − 5)]′ = cos(x − 5) · (x − 5)′ = cos(x − 5) · (x′ − 5′ ) = cos(x − 5) · 1 = cos(x − 5)
3. [sin(4x + 3)]′ = cos(4x + 3) · (4x + 3)′ = cos(4x + 3) · 4 = 4 cos(4x + 3)
4. [cos(2 − x)]′ = − sin(2 − x) · (2 − x)′ = − sin(2 − x) · (−1) = sin(2 − x)
5. [(3x2 − 5x + 1)50 ]′ =
2
Section 2.5 The Chain Rule
2010 Kiryl Tsishchanka
DIFFERENTIATION RULES
c′ = 0
(un )′ = nun−1 · u′
[cf (x)]′ = cf ′ (x)
(sin u)′ = cos u · u′
(csc u)′ = − csc u cot u · u′
[f (x) ± g(x)]′ = f ′ (x) ± g ′ (x)
(cos u)′ = − sin u · u′ (sec u)′ = sec u tan u · u′
2
(tan u) = sec u · u
′
2
(cot u) = − csc u · u
′
′
′
[f (x)g(x)]′ = f ′ (x)g(x) + f (x)g ′ (x)
f (x)
g(x)
′
=
f ′ (x)g(x) − f (x)g ′ (x)
[g(x)]2
EXAMPLES:
5.
[(3x2 − 5x + 1)50 ]′ = 50(3x2 − 5x + 1)50−1 · (3x2 − 5x + 1)′ = 50(3x2 − 5x + 1)49 (6x − 5)
6.
√
1
1
8
[ 3 1 − 4x2 ]′ = (1 − 4x2 )1/3−1 · (1 − 4x2 )′ = (1 − 4x2 )−2/3 · (−8x) = − x(1 − 4x2 )−2/3
3
3
3
7.
[tan(x3 )]′ = sec2 (x3 ) · (x3 )′ = sec2 (x3 ) · (3x2 ) = 3x2 sec2 (x3 )
8.
[tan3 x]′ = 3 tan2 x · (tan x)′ = 3 tan2 x sec2 x
9.
[tan3 (x3 )]′ = 3 tan2 (x3 ) · [tan(x3 )]′ = [by (7)] = 3 tan2 (x3 ) · 3x2 sec2 (x3 )
= 9x2 tan2 (x3 ) sec2 (x3 )
10.
3
tan x
√
x
′
3
=
[tan x]
′
√
√
1
3 tan2 x sec2 x x − tan3 x · √
x − tan x( x)
2 x
√ 2
= [by (8)] =
x
( x)
3
√
′
11. [5x + 7(1 + 2x)10 ]′ = 5x′ + 7[(1 + 2x)10 ]′ = 5 · 1 + 7 · 10(1 + 2x)10−1 · (1 + 2x)′
= 5 + 70(1 + 2x)9 · 2
= 5 + 140(1 + 2x)9
12. [x sec(1 − x)]′ = x′ sec(1 − x) + x[sec(1 − x)]′ = sec(1 − x) + x sec(1 − x) tan(1 − x) · (1 − x)′
= sec(1 − x) + x sec(1 − x) tan(1 − x) · (−1)
= sec(1 − x) − x sec(1 − x) tan(1 − x)
= sec(1 − x)[1 − x tan(1 − x)]
13. [(2x + 1)2 cos(5 − 3x)]′ =
3
Section 2.5 The Chain Rule
2010 Kiryl Tsishchanka
DIFFERENTIATION RULES
c′ = 0
(un )′ = nun−1 · u′
[cf (x)]′ = cf ′ (x)
(sin u)′ = cos u · u′
(csc u)′ = − csc u cot u · u′
[f (x) ± g(x)]′ = f ′ (x) ± g ′ (x)
(cos u)′ = − sin u · u′ (sec u)′ = sec u tan u · u′
[f (x)g(x)]′ = f ′ (x)g(x) + f (x)g ′ (x)
(tan u)′ = sec2 u · u′
(cot u)′ = − csc2 u · u′
f (x)
g(x)
′
=
f ′ (x)g(x) − f (x)g ′ (x)
[g(x)]2
EXAMPLES:
13. [(2x + 1)2 cos(5 − 3x)]′ = [(2x + 1)2 ]′ cos(5 − 3x) + (2x + 1)2 [cos(5 − 3x)]′
= 2(2x + 1) · (2x + 1)′ cos(5 − 3x) + (2x + 1)2 (− sin(5 − 3x))(5 − 3x)′
= 2(2x + 1) · 2 · cos(5 − 3x) + (2x + 1)2 (− sin(5 − 3x)) · (−3)
= 4(2x + 1) cos(5 − 3x) + 3(2x + 1)2 sin(5 − 3x)
14.
"s
4
#′
√
√
√
1/4−1 ′
(1 − x) 3x − 8
1 (1 − x) 3x − 8
(1 − x) 3x − 8
=
·
4
sin2 (1 − 5x)
sin2 (1 − 5x)
sin2 (1 − 5x)
√
−3/4
(1 − x) 3x − 8
sin2 (1 − 5x)
√
√
[(1 − x) 3x − 8]′ · sin2 (1 − 5x) − (1 − x) 3x − 8 · [sin2 (1 − 5x)]′
×
sin4 (1 − 5x)
1
=
4
√
−3/4
(1 − x) 3x − 8
sin2 (1 − 5x)
√
√
√
[(1 − x)′ 3x − 8 + (1 − x)( 3x − 8)′ ] · sin2 (1 − 5x) − (1 − x) 3x − 8 · [sin2 (1 − 5x)]′
×
sin4 (1 − 5x)
1
=
4
√
−3/4
(1 − x) 3x − 8
sin2 (1 − 5x)
√
√
(− 3x − 8 + 23 (1 − x)(3x − 8)−1/2 ) sin2 (1 − 5x) + 10(1 − x) 3x − 8 sin(1 − 5x) cos(1 − 5x)
×
sin4 (1 − 5x)
1
=
4
4
Section 2.5 The Chain Rule
2010 Kiryl Tsishchanka
EXERCISES:
1. [sin(3x)]′ =
2. [4 cos(x3 )]′ =
3. [x(x2 − x + 1)23 ]′ =
√
4. [ x3 + csc x]′ =
5.
1
3
x + 2x − 3
6. [sec
√
′
=
1 + cos x]′ =
5
Section 2.5 The Chain Rule
2010 Kiryl Tsishchanka
SOLUTIONS:
1. [sin(3x)]′ = cos(3x) · (3x)′ = cos(3x) · 3 = 3 cos(3x)
2. [4 cos(x3 )]′ = 4[cos(x3 )]′ = −4 sin(x3 ) · (x3 )′ = −4 sin(x3 ) · (3x2 ) = −12x2 sin(x3 )
3. [x(x2 − x + 1)23 ]′ = x′ (x2 − x + 1)23 + x[(x2 − x + 1)23 ]′
= (x2 − x + 1)23 + x · 23(x2 − x + 1)23−1 · (x2 − x + 1)′
= (x2 − x + 1)23 + 23x(x2 − x + 1)22 (2x − 1)
√
1
4. [ x3 + csc x]′ = [(x3 + csc x)1/2 ]′ = (x3 + csc x)1/2−1 (x3 + csc x)′
2
1
= (x3 + csc x)−1/2 (3x2 − csc x cot x)
2
5.
1
x3 + 2x − 3
6. [sec
√
′
= [(x3 + 2x − 3)−1 ]′ = (−1)(x3 + 2x − 3)−1−1 (x3 + 2x − 3)′
= −(x3 + 2x − 3)−2 (3x2 + 2)
√
√
√
1 + cos x]′ = sec 1 + cos x tan 1 + cos x [ 1 + cos x]′
√
√
1
= sec 1 + cos x tan 1 + cos x (1 + cos x)−1/2 (1 + cos x)′
2
√
√
1
= sec 1 + cos x tan 1 + cos x (1 + cos x)−1/2 (− sin x)
2
√
√
1
= − sec 1 + cos x tan 1 + cos x (1 + cos x)−1/2 sin x
2
COMMON MISTAKES
1. [(1 − x)3 ]′ = 3(1 − x)2 WRONG!!!
Solution: By the Chain Rule we have:
[(1 − x)3 ]′ = 3(1 − x)2 · (1 − x)′ = 3(1 − x)2 · (−1) = −3(1 − x)2
√
2. [sin( x)] = cos
′
1
√
WRONG!!!
2 x
Solution: By the Chain Rule we have
√
√
√
√
1
[sin( x)]′ = cos( x) · ( x)′ = cos( x) · √
2 x
6
Section 2.5 The Chain Rule
2010 Kiryl Tsishchanka
Appendix
EXAMPLE: Let f (x) =
√
1 + x3 . Find f ′ , f ′′ , and f ′′′ .
Solution: Since f (x) = (1 + x3 )1/2 , we have
3x2
1
1
f ′ (x) = (1 + x3 )1/2−1 · (1 + x3 )′ = (1 + x3 )−1/2 · 3x2 = √
2
2
2 1 + x3
′′
f (x) =
3x2
2(1 + x3 )1/2
′
3
=
2
=
x2
(1 + x3 )1/2
′
3 (x2 )′ (1 + x3 )1/2 − x2 [(1 + x3 )1/2 ]′
·
2
[(1 + x3 )1/2 ]2
3 1/2−1
3 1/2
21
· (1 + x3 )′
3 2x(1 + x ) − x 2 (1 + x )
= ·
2
1 + x3
3 1/2
21
(1 + x3 )−1/2 · 3x2
2x(1
+
x
)
−
x
3
2
= ·
2
1 + x3
3 4
3 −1/2
3 1/2
3 2x(1 + x ) − 2 x (1 + x )
= ·
2
1 + x3
3 4
3 −1/2
3 1/2
· 2(1 + x3 )1/2
2x(1 + x ) − x (1 + x )
3
2
= ·
2
(1 + x3 ) · 2(1 + x3 )1/2
3 4
3 −1/2
3 1/2
3 1/2
· 2(1 + x3 )1/2
3 2x(1 + x ) · 2(1 + x ) − 2 x (1 + x )
= ·
2
2(1 + x3 )3/2
=
3 4x(1 + x3 ) − 3x4
·
2
2(1 + x3 )3/2
=
3 4x + 4x4 − 3x4
·
2
2(1 + x3 )3/2
=
3
4x + x4
·
2 2(1 + x3 )3/2
=
3 x(4 + x3 )
·
2 2(1 + x3 )3/2
3x(4 + x3 )
=
4(1 + x3 )3/2
7
Section 2.5 The Chain Rule
2010 Kiryl Tsishchanka
′
3x(4 + x3 )
f (x) =
4(1 + x3 )3/2
′
3 x(4 + x3 )
=
4 (1 + x3 )3/2
′′′
=
3 [x(4 + x3 )]′ (1 + x3 )3/2 − x(4 + x3 )[(1 + x3 )3/2 ]′
·
4
[(1 + x3 )3/2 ]2
3 3/2−1
′
3
3 ′
3 3/2
3 3
· (1 + x3 )′
3 [x (4 + x ) + x(4 + x ) ](1 + x ) − x(4 + x ) 2 (1 + x )
= ·
4
(1 + x3 )3
3
2
3 3/2
3 3
[1
·
(4
+
x
)
+
x
·
3x
](1
+
x
)
−
x(4
+
x
) (1 + x3 )1/2 · 3x2
3
2
= ·
3
3
4
(1 + x )
9 3
3
3 1/2
3
3
3 3/2
3 (4 + x + 3x )(1 + x ) − 2 x (4 + x )(1 + x )
= ·
4
(1 + x3 )3
9 3
3
3 1/2
3
3 3/2
3 (4 + 4x )(1 + x ) − 2 x (4 + x )(1 + x )
= ·
4
(1 + x3 )3
9 3
3
3 1/2
3
3 3/2
3 4(1 + x )(1 + x ) − 2 x (4 + x )(1 + x )
= ·
4
(1 + x3 )3
9 3
3 5/2
x (4 + x3 )(1 + x3 )1/2
4(1
+
x
)
−
3
2
= ·
4
(1 + x3 )3
9 3
3
3 1/2
3 5/2
4(1 + x ) − x (4 + x )(1 + x )
· 2(1 + x3 )−1/2
3
2
= ·
4
(1 + x3 )3 · 2(1 + x3 )−1/2
9
3 5/2
3 −1/2
− x3 (4 + x3 )(1 + x3 )1/2 · 2(1 + x3 )−1/2
3 4(1 + x ) · 2(1 + x )
2
= ·
4
2(1 + x3 )5/2
=
3 8(1 + 2x3 + x6 ) − 36x3 − 9x6
3 8(1 + x3 )2 − 9x3 (4 + x3 )
·
·
=
4
2(1 + x3 )5/2
4
2(1 + x3 )5/2
=
3 8 − 20x3 − x6
3 8 + 16x3 + 8x6 − 36x3 − 9x6
·
=
·
4
2(1 + x3 )5/2
4 2(1 + x3 )5/2
=
3(8 − 20x3 − x6 )
3(x6 + 20x3 − 8)
=
−
8(1 + x3 )5/2
8(1 + x3 )5/2
8