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 calculusrules:

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)21 = 2(1 + x)

Note that in both cases we got the same result. However, the goal of Section 2.5 is to showthat despite the fact that Solution 2 gives the right answer, it is not completely correct. Toexplain what me mean by that, let us consider the following example:

PROBLEM: Let f(x) = (1 x)2. Find f (x).Solution 1: We have

f (x) = [(1 x)2] = (1 2x + x2) = 1 2(x) + (x2)

= 0 2 1 + 2x = 2 + 2x

Solution 2(???): If we apply the power rule immediately,we get

f (x) = [(1 x)2] ?= 2(1 x)21 = 2(1 x)

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 ofthe tangent line to the curve f(x) = (1 x)2 at x = 2 is negative. But this is not the case!

x

4

y

3

8

6

2

4

2

10

-2

0-1-2

CONCLUSION: We cant always apply the rules (xn) = nxn1, (sin x) = cosx, etc. to caseswhen 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 [(2x)3] 6= 3(2x)2, [(2 x)3] 6= 3(2 x)2

(sin x) = cosx, [sin(x 5)] = cos(x 5) [sin(4x)] 6= cos(4x), [sin(5 x)] 6= cos(5 x)

(

x) =1

2

x, (

x 3) = 1

2

x 3 (5x) 6= 1

25x

, (3 x) 6= 1

23 x

1

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 functiondefined 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

dx=

dy

du

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((1 x)2)dx

= [y = u2, u = 1 x] = d(u2)

du

d(1 x)dx

= 2u (1) = 2(1 x)(1) = 2(1 x)

DIFFERENTIATION RULES

c = 0 (un) = nun1 u [cf(x)] = cf (x)

(sin u) = cosu u (csc u) = csc u cotu u [f(x) g(x)] = f (x) g(x)

(cosu) = sin u u (sec u) = sec u tanu u [f(x)g(x)] = f (x)g(x) + f(x)g(x)

(tanu) = sec2 u u (cotu) = csc2 u u[f(x)

g(x)

]

=f (x)g(x) f(x)g(x)

[g(x)]2

EXAMPLES:

1. [sin x] = cosx x = cosx 1 = cosx

2. [sin(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) = nun1 u [cf(x)] = cf (x)

(sin u) = cosu u (csc u) = csc u cotu u [f(x) g(x)] = f (x) g(x)

(cosu) = sin u u (sec u) = sec u tanu u [f(x)g(x)] = f (x)g(x) + f(x)g(x)

(tanu) = sec2 u u (cotu) = csc2 u u[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)49 (3x2 5x + 1) = 50(3x2 5x + 1)49(6x 5)

6. [ 31 4x2] = 1

3(1 4x2)2/3 (1 4x2) = 1

3(1 4x2)2/3 (8x) = 8

3x(1 4x2)2/3

7. [tan(x3)] = sec2(x3) (x3) = sec2(x3) (3x2) = 3x2 sec2(x3)

8. [tan3 x] = 3 tan2 x (tanx) = 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.

[tan3 x

x

]

=[tan3 x]

x tan3 x(x)(

x)2= [by (8)] =

3 tan2 x sec2 x

x tan3 x 12

x

x

11. [5x + 7(1 + 2x)10] = 5x + 7[(1 + 2x)10] = 5 1 + 7 10(1 + 2x)9 (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) = nun1 u [cf(x)] = cf (x)

(sin u) = cosu u (csc u) = csc u cotu u [f(x) g(x)] = f (x) g(x)

(cosu) = sin u u (sec u) = sec u tanu u [f(x)g(x)] = f (x)g(x) + f(x)g(x)

(tanu) = sec2 u u (cotu) = 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.

[4

(1 x)3x 8sin2(1 5x)

]=

1

4

((1 x)3x 8sin2(1 5x)

)3/4[(1 x)3x 8sin2(1 5x)

]

=1

4

((1 x)3x 8sin2(1 5x)

)3/4

[(1 x)3x 8] sin2(1 5x) (1 x)3x 8 [sin2(1 5x)]

sin4(1 5x)

=1

4

((1 x)3x 8sin2(1 5x)

)3/4

((1 x)3x 8 + (1 x)[3x 8]) sin2(1 5x) (1 x)3x 8 [sin2(1 5x)]

sin4(1 5x)

=1

4

((1 x)3x 8sin2(1 5x)

)3/4

(3x 8 + 3

2(1 x)(3x 8)1/2) sin2(1 5x) + 10(1 x)3x 8 sin(1 5x) cos(1 5x)

sin4(1 5x)

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

x3 + 2x 3

]

=

6. [sec1 + cosx] =

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)22 (x2 x + 1)

= (x2 x + 1)23 + 23x(x2 x + 1)22(2x 1)

4. [

x3 + csc x] = [(x3 + csc x)1/2] =1

2(x3 + csc x)1/2(x3 + csc x)

=1

2(x3 + csc x)1/2(3x2 csc x cotx)

5.

[1

x3 + 2x 3]

= [(x3 + 2x 3)1] = (1)(x3 + 2x 3)2(x3 + 2x 3)

= (x3 + 2x 3)2(3x2 + 2)

6. [sec1 + cos x] = sec

1 + cosx tan

1 + cos x [

1 + cosx]

= sec1 + cosx tan

1 + cos x

1

2(1 + cos x)1/2(1 + cosx)

= sec1 + cosx tan

1 + cos x

1

2(1 + cos x)1/2( sin x)

= 12sec

1 + cosx tan

1 + cosx (1 + cos x)1/2 sin x

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

2

x

)WRONG!!!

Solution: By the Chain Rule we have

[sin(

x)] = cos(

x) (x) = cos(x) 12

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

f (x) =1

2(1 + x3)1/21 (1 + x3) = 1

2(1 + x3)1/2 3x2 = 3x

2

21 + x3

f (x) =

(3x2

2(1 + x3)1/2

)

=3

2

(x2

(1 + x3)1/2

)

=3

2 (x

2)(1 + x3)1/2 x2[(1 + x3)1/2][(1 + x3)1/2]2

=3

22x(1 + x3)1/2 x2 1

2(1 + x3)1/21 (1 + x3)

1 + x3

=3

22x(1 + x3)1/2 x2 1

2(1 + x3)1/2 3x2

1 + x3

=3

22x(1 + x3)1/2 3

2x4(1 + x3)1/2

1 + x3

=3

2

(2x(1 + x3)1/2 3

2x4(1 + x3)1/2

) 2(1 + x3)1/2

(1 + x3) 2(1 + x3)1/2

=3

22x(1 + x3)1/2 2(1 + x3)1/2 3

2x4(1 + x3)1/2 2(1 + x3)1/2

2(1 + x3)3/2

=3

2 4x(1 + x

3) 3x42(1 + x3)3/2

=3

2 4x + 4x

4 3x42(1 + x3)3/2

=3

2 4x + x

4

2(1 + x3)3/2

=3

2 x(4 + x

3)

2(1 + x3)3/2

=3x(4 + x3)

4(1 + x3)3/2

7

Section 2.5 The Chain Rule 2010 Kiryl Tsishchanka

f (x) =

(3x(4 + x3)

4(1 + x3)3/2

)

=3

4

(x(4 + x3)

(1 + x3)3/2

)

=3

4 [x(4 + x

3)](1 + x3)3/2 x(4 + x3)[(1 + x3)3/2][(1 + x3)3/2]2

=3

4[x(4 + x3) + x(4 + x3)](1 + x3)3/2 x(4 + x3)3

2(1 + x3)3/21 (1 + x3)

(1 + x3)3

=3

4[1 (4 + x3) + x 3x2](1 + x3)3/2 x(4 + x3)3

2(1 + x3)1/2 3x2

(1 + x3)3

=3

4(4 + x3 + 3x3)(1 + x3)3/2 9

2x3(4 + x3)(1 + x3)1/2

(1 + x3)3

=3

4(4 + 4x3)(1 + x3)3/2 9

2x3(4 + x3)(1 + x3)1/2

(1 + x3)3

=3

44(1 + x3)(1 + x3)3/2 9

2x3(4 + x3)(1 + x3)1/2

(1 + x3)3

=3

44(1 + x3)5/2 9

2x3(4 + x3)(1 + x3)1/2

(1 + x3)3

=3

4

(4(1 + x3)5/2 9

2x3(4 + x3)(1 + x3)1/2

) 2(1 + x3)1/2

(1 + x3)3 2(1 + x3)1/2

=3

44(1 + x3)5/2 2(1 + x3)1/2 9

2x3(4 + x3)(1 + x3)1/2 2(1 + x3)1/2

2(1 + x3)5/2

=3

4 8(1 + x

3)2 9x3(4 + x3)2(1 + x3)5/2

=3

4 8(1 + 2x

3 + x6) 36x3 9x62(1 + x3)5/2

=3

4 8 + 16x

3 + 8x6 36x3 9x62(1 + x3)5/2

=3

4 8 20x

3 x62(1 + x3)5/2

=3(8 20x3 x6)8(1 + x3)5/2

= 3(x6 + 20x3 8)8(1 + x3)5/2

8