The Chain Rule

Section 4.3 Problems

4.4 The Chain Rule and Higher Derivatives

4.4.1 The Chain Rule

In Section 1.2, we defined the composition of functions. To find the derivative of composite functions, we need thechain rule, the proof of which is given at the end of this section.

Chain Rule Ifgis differentiable atx and f is differentiable aty=g(x)^{, then} the composite function(f ◦g)(x) = f[g(x)]is differentiable atx, and the derivative is given by

(f ◦g)(x)= f[g(x)]g(x)

This formula looks complicated. Let’s take a moment to see what we need to do to find the derivative of the composite function(f ◦g)(x). The functiongis the inner function; the function f is the outer function. The expression f[g(x)]g(x) thus means that we need to find the derivative of the outer function, evaluated at g(x), and the derivative of the inner function, evaluated atx, and then multiply the two together.

EXAMPLE 1 A Polynomial Find the derivative of

h(x)=(³x²−¹)²

Solution

The inner function isg(x)=3x²−1; the outer function is f(u)=u². Then g(x)=⁶x and f(u)=²u

Evaluating f(u)^atu=g(x)^yields

f[g(x)] =2g(x)=2(3x²−1) Thus,

h(x)=(f ◦g)(x)= f[g(x)]g(x)

=²(³x²−¹)⁶x =¹²x(³x²−¹)

The derivative of f ◦gcan be written in Leibniz notation. If we setu=g(x)^{, then} d

d x ^[(f ◦g)(x)^]= d f du

du d x

This form of the chain rule emphasizes that, in order to differentiate f◦g, we multiply the derivative of the outer function and the derivative of the inner function, the former evaluated atu, the latter atx.

EXAMPLE 2 A Polynomial Find the derivative of

h(x)=(²x+¹)³

Solution

If we setu=g(x)=²x+^{1 and} f(u)=u³, thenh(x)=(f ◦g)(x). We need to find both f[g(x)]^andg(x)^{to compute}h(x)^{. Now,}

g(x)=^{2 and} f(u)=³u² Hence, since f[g(x)] =³(g(x))² =³(²x+¹)², it follows that

h(x)= f[g(x)]g(x)=³(²x+¹)²·²

=⁶(²x+¹)² If we use Leibniz notation, this becomes

h(x)= d f du

du

d x =3u²·2=3(2x+1)²·2

=⁶(²x+¹)²

EXAMPLE 3 A Radical Find the derivative ofh(x)= x²+^1.

Solution

If we setu=g(x)=x²+^{1 and} f(u)=√

u, thenh(x)=(f ◦g)(x). We find that g(x)=²x and f(u)= ¹

2√ u We need to evaluate fatg(x)—that is,

f[g(x)] = ¹ 2√

g(x) = ¹ 2

x²+¹ Therefore,

h(x)= f[g(x)]g(x)= ¹ 2

x²+¹²x = x x²+¹

EXAMPLE 4 A Radical Find the derivative of

h(x)=⁷

2x²+³x

Solution

We write

h(x)=(²x²+³x)^1/7

The inner function isu = g(x) = ²x²+³x and the outer function is f(u) = u^1/7. Thus, we find that

h(x)= d f du

du d x = ¹

7u^1/7−1(⁴x+³)

= ¹

7(2x²+3x)^−6/7(4x+3)

= ⁴x+³ 7(²x²+³x)^6/7

EXAMPLE 5 A Rational Function Find the derivative ofh(x)=

x+1x

2

.

Solution

If we setu=g(x)= _x+^x₁and f(u)=u², thenh(x)=(f◦g)(x). We use the quotient rule to compute the derivative ofg(x)^:

g(x)= ¹·(x+¹)−x·¹

(x+¹)² = ¹

(x+¹)² Since f(u)=²u, we obtain

h(x)= f[g(x)]g(x)=² x x+¹

1

(x+¹)² = ²x (x+¹)³

The Proof of the Quotient Rule

We can use the chain rule to prove the quotient rule. Assume thatg(x)=^{0 for all}xin the domain ofg. If we defineh(x)= ¹_x^{, then}

(h◦g)(x)=h[g(x)] = ¹ g(x)

We used the formal definition of the derivative in Example 3 in Section 4.1 to show thath(x)= −_x¹2. This, together with the chain rule, yields

(h◦g)(x)= − ¹

[g(x)]²g(x), or

1 g

= −g g²

Since _g^f = f ¹_g, we can use the product rule to find the derivative of _g^f: f

g

= f¹ g + f

1 g

= f¹ g + f

−g g²

= fg− f g g²

Note that we did not use the power rule for negative integer exponents (Subsection 4.3.2) to compute h(x), but instead used the formal definition of derivatives to compute the derivative of 1/x. Using the power rule for negative integer exponents would have been circular reasoning: We used the quotient rule to prove the power rule for negative integer exponents, so we cannot use the power rule for negative integer exponents to prove the quotient rule.

EXAMPLE 6 A Function with Parameters Find the derivative of h(x)=(ax²−²)ⁿ wherea>^{0 and}nis a positive integer.

Solution

If we setu=g(x)=ax²−^{2 and} f(u)=uⁿ, thenh(x)=(f ◦g)(x)^{. Since} g(x)=²ax and f(u)=nuⁿ⁻¹

it follows that

h(x)= f[g(x)]g(x)=n(ax²−2)ⁿ⁻¹2ax

=²anx(ax²−²)ⁿ⁻¹

Looking ath(x)=n(ax²−²)ⁿ⁻¹·²ax, we see that we first differentiated the outer function f, which yieldedn(ax²−²)ⁿ⁻¹via the power rule, and then multiplied the result by 2ax, the derivative of the inner functiong.

EXAMPLE 7 Differentiating a Function That Is Not Specified Suppose f(x)is differentiable. Find d

d x

√ 1 f(x)

Solution

We set

h(x)= √¹

f(x) = [f(x)]^−1/2

Now,u= f(x)is the inner function andh(u)=u^−1/2is the outer function; hence, d

d xh(x)= dh du

du d x = −¹

2u^−3/2f(x)

= − ¹

2u^3/2 f(x)= − f(x) 2[f(x)]^3/2

EXAMPLE 8 Generalized Power Rule Suppose f(x)is differentiable andris a real number. Find d

d x[f(x)]^r

Solution

Using the general form of the power rule and the chain rule, we find that d

d x[f(x)]^r =r[f(x)]^r−1f(x)

EXAMPLE 9 Differentiating a Function That Is Not Specified Suppose that f(x)=³x−^{1. Find} d

d x f(x²) ^atx =³

Solution

The inner function isu=x², the outer function is f(u), and we find that d

d x f(x²)=²x f(x²)

If we substitutex =^{3 into} f(x²), we obtain f(³²)= f(⁹)=(³)(⁹)−¹=^{26. Thus,} d

d x f(x²)

x=3=(²)(³)f(⁹)=(⁶)(²⁶)=¹⁵⁶

The chain rule can be applied repeatedly, as shown in the next two examples.

EXAMPLE 10 Nested Chain Rule Find the derivative of h(x)=

x²+¹+¹ 2

Solution

If we seth(x)= (f ◦g)(x), theng(x)=

x²+1+1 and f(u)=u². We see that g(x)is itself a composition of two functions, with inner functionv=x²+1 and outer function√

v+1. To differentiateh(x), we proceed stepwise. First,

h(x)= d d x

x²+¹+¹ 2

=²

x²+¹+¹ d

d x

x²+¹+¹

Then, since

d d x

x²+¹+¹

= ²x 2

x²+¹ = x x²+¹ (where we used the chain rule to differentiate

x²+^{1), we get} h(x)=²

x²+¹+¹ x

x²+¹

EXAMPLE 11 Nested Chain Rule Find the derivative of h(x)=

2x³− 3x⁴−2

3

Solution

As in the previous example, we proceed stepwise:

h(x)=³

2x³− 3x⁴−²

2

d d x

2x³− 3x⁴−²

=³

2x³− 3x⁴−²

2

6x²− ¹²x³ 2

3x⁴−²

=¹⁸x²

2x³− 3x⁴−²

2

1− x

3x⁴−²

We conclude this subsection with the proof of the chain rule. The first part of the proof follows along the lines of the argument we sketched out at the beginning of the section, but the second part is much more technical and deals with the problem that ucould be zero.

Proof of the Chain Rule

We will use the definition of the derivative to prove the chain rule. Formally,

(f ◦g)(x)= ^lim

x→c

f[g(x)] − f[g(c)]

x−c

We need to show that the right-hand side is equal to f[g(c)]g(c). As long asg(x)= g(c), we can write

x→clim

f[g(x)] − f[g(c)]

x−c = ^lim

x→c

f[g(x)]−f[g(c)]

g(x)−g(c) [g(x)−g(c)]

x−c

= ^lim

x→c

f[g(x)] − f[g(c)]

g(x)−g(c)

g(x)−g(c) x−c

Sinceg(x)is continuous atx =c, it follows that lim_x→cg(x)=g(c), and hence,

x→clim

f[g(x)] − f[g(c)]

g(x)−g(c) = f[g(c)]

Furthermore,

x→clim

g(x)−g(c)

x−c =g(c)

Since these limits exist, we can use the fact that the limit of a product is the product of the limits. We find that

x→clim

f[g(x)] − f[g(c)]

x−c = ^lim

x→c

f[g(x)] − f[g(c)]

g(x)−g(c) x→c^lim

g(x)−g(c) x−c

= f[g(c)]g(c)

In the preceding calculation, we needed to assume thatg(x)−g(c) = ^{0. Of} course, when we take the limit asx → c, there might bex-values such thatg(x)= g(c), and we must deal with this possibility.

We sety=g(x)^andd=g(c). The expression f^∗(y)≡ f(y)− f(d)

y−d is defined only fory =d. Since

y→dlim

f(y)− f(d)

y−d = f(d)

we can extend f^∗[g(x)] ^{by defining} f^∗[g(x)] = f[g(x)] ^{to make} f^∗[g(x)] ^a continuous function:

f^∗[g(x)] =

⎧⎪

⎨

⎪⎩

f[g(x)] − f[g(c)]

g(x)−g(c) ^forg(x)=g(c) f[g(c)] forg(x)=g(c) This means that, for allx,

f[g(x)] − f[g(c)] = f^∗[g(x)][g(x)−g(c)]

With this equivalence, we can repeat our calculations to obtain

x→clim

f[g(x)] − f[g(c)]

x−c = ^lim

x→c

f^∗[g(x)][g(x)−g(c)]

x−c

x→clim f^∗[g(x)]^lim

x→c

g(x)−g(c)

x−c = f[g(c)] ·g(c)

Note that in the last step we used the fact that f^∗[g(x)]is continuous atx=c.