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# 2.5: The Chain Rule

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## Learning Objectives

A student will be able to:

• Know the chain rule and its proof.
• Apply the chain rule to the calculation of the derivative of a variety of composite functions.

We want to derive a rule for the derivative of a composite function of the form $f \circ g$ in terms of the derivatives of $f$ and $g$. This rule allows us to differentiate complicated functions in terms of known derivatives of simpler functions.

The Chain Rule

If $g$ is a differentiable function at $x$ and $f$ is differentiable at $g(x)$, then the composition function $f \circ g = f(g(x))$ is differentiable at $x$. The derivative of the composite function is:

$(f \circ g)'(x) = f '(g(x))g '(x).$

Another way of expressing, if $u = u(x)$ and $f = f(u)$, then

$\frac {d}{dx}[f(u)] = f'(u)\frac {du}{dx}.$

And a final way of expressing the chain rule is the easiest form to remember: If $y$ is a function of $u$ and $u$ is a function of $x$, then

$\frac {dy}{dx}= \frac {dy}{du}.\frac {du}{dx}.$

Example 1:

Differentiate $f(x) = (2x^3 - 4x^2 + 5)^2.$

Solution:

Using the chain rule, let $u = 2x^3 - 4x^2 + 5.$ Then

$\frac {d}{dx}[(2x^2-4x^2+5)^2] & = \frac {d}{dx}[u^2]\\& = 2u \frac {du}{dx}\\& = 2(2x^3-4x^2+5)(6x^2-8x).$

The example above is one of the most common types of composite functions. It is a power function of the type

$y = [u(x)]^n.$

The rule for differentiating such functions is called the General Power Rule. It is a special case of the Chain Rule.

The General Power Rule

if

$y = \left[ u(x) \right] ^n$

then

$\frac {dy}{dx} = n[u(x)]^{n-1} u'(x).$

In simpler form, if

$y = u^n$

then

$y' = n u^{n-1} \cdot u'.$

Example 2:

What is the slope of the tangent line to the function $y = \sqrt {x^2 - 3x + 2}$ that passes through point $x = 3$?

Solution:

We can write $y = (x^2 - 3x + 2)^{1/2}.$ This example illustrates the point that $n$ can be any real number including fractions. Using the General Power Rule,

$\frac{dy} {dx} &= \frac{1} {2} (x^2 - 3x + 2)^{\frac{1} {2} - 1}(2x - 3)\\&= \frac{1} {2} (x^2 - 3x + 2)^{-1/2}(2x - 3)\\&= \frac{(2x - 3)} {2 \sqrt{x^2 - 3x + 2}}$

To find the slope of the tangent line, we simply substitute $x = 3$ into the derivative:

$\frac{dy} {dx}\Bigg|_{x = 3}= \frac{2(3) - 3} {2 \sqrt{3^2 - 3(3) + 2}} = \frac{3} {2 \sqrt{2}} = \frac{3\sqrt{2}} {4}.$

Example 3:

Find $dy/dx$ for $y = \sin^3x$.

Solution:

The function can be written as $y = [\sin x]^3.$ Thus

$\frac{dy} {dx} &= 3[\sin x]^2 [\cos x] \\&= 3 \sin^{2} x \cos x$

Example 4:

Find $dy/dx$ for $y = 5 \cos(3x^2 - 1).$

Solution:

Let $u = 3x^2 - 1.$ By the chain rule,

$\frac{d} {dx} [f(u)] = f'(u) \frac{du} {dx}$

where $f(u) = 5 \cos u.$ Thus

$\frac{dy} {dx} & = 5(-\sin u)(6x)\\& = -5(6x)\sin u\\& = -30x \sin(3x^2 - 1)$

Example 5:

Find $dy/dx$ for $y= [\cos(\pi x^2)]^3.$

Solution:

This example applies the chain rule twice because there are several functions embedded within each other.

Let $u$ be the inner function and $w$ be the innermost function.

$y = (u(w))^3\\ u(x) = \cos x \\w (x)= \pi x^2.$

Using the chain rule,

$\frac{d} {dx} [f(u)] &= f'(u) \frac{du} {dx}\\ \frac{d} {dx} [u^3] &= \frac{d} {dx} [\cos^3(\pi x^2)]\\&= \frac{d} {dx} [\cos(\pi x^2)]^3\\&= 3[\cos(\pi x^2)]^2 [-\sin(\pi x^2)](2\pi x)\\&= -6\pi x [\cos(\pi x^2)]^2 \sin(\pi x^2).$

Notice that we used the General Power Rule and, in the last step, we took the derivative of the argument.

For an introduction to the Chain Rule (5.0), see Khan Academy, Calculus: Derivatives 4: The Chain Rule (9:11).

For more examples of the Chain Rule (5.0), see Math Video Tutorials by James Sousa The Chain Rule: Part 1 of 2 (8:45).

## Review Questions

Find $f'(x)$.

1. $f(x) = (2x^2 - 3x)^{39}$
2. $f(x) = \left (x^3 - \frac{5} {x^2} \right)^{-3}$
3. $f(x) = \frac{1} {\sqrt{3x^2 -6x + 2}}$
4. $f(x) = \sin^3x$
5. $f(x) = \sin x^3$
6. $f(x) = \sin^3x^3$
7. $f(x) = \tan(4x^5)$
8. $f(x) = \sqrt{4x - \sin^2 2x}$
9. $f(x) = \frac{\sin x} {\cos(3x - 2)}$
10. $f(x) = (5x + 8)^3 (x^3 + 7x)^{13}$
11. $f(x) = \left (\frac{x - 3} {2x - 5} \right)^3$

1. $f'(x) = 39(2x^2 - 3x)^{38} (4x - 3)$
2. $f'(x) = -3 \left (\frac{x^2} {x^5 - 5} \right)^4 \left(\frac{3x^5 - 10} {x^3} \right)$
3. $f'(x) = \frac{-3(x - 1)} {\sqrt{(3x^2 -6x + 1)}^3}$
4. $f'(x) = 3 \sin^2 x \cos x$
5. $f'(x) = 3x^2 \cos x^3$
6. $f'(x) = 9x^2 \cos x^3 \sin^2 x^3$
7. $f'(x) = 20x^4 \sec^2(4x^5)$
8. $f'(x) = \frac{2(1 - \sin 2x \\cos 2x)} {\sqrt{4x - \sin^2 2x}}$
9. $f'(x) = \frac{\cos(3x - 2) \cos x + 3 \sin(3x - 2) \sin x} {\cos^2(3x - 2)}$
10. $f'(x) = 13(5x + 8)^3 (x^3 + 7x)^{12} (3x^2 + 7) + 15(x^3 + 7x)^{13} (5x + 8)^2$
11. To find $\frac{df}{dx}$, let’s start by defining $u(x) = \frac{x - 3}{2x - 5}$. Then, by the powers rule: $\frac{df}{du} = \frac{du^3}{du} = 3u^2 = 3 \left (\frac{x - 3}{2x - 5} \right )^2.$ Also, by the quotient rule: $\frac{du}{dx} = \frac{(2x - 5)(1) - (x - 3)(2)}{(2x - 5)^2} = \frac{2x - 5 - 2x + 6}{(2x - 5)^2} = \frac{1}{(2x - 5)^2}.$ Finally, the chain rule gives us: $\frac{df}{dx} = \frac{df}{du} \frac{du}{dx} = \frac{3x^2 - 18x + 27}{(2x - 5)^4}$

Feb 23, 2012

Sep 01, 2014