5.5: Chain Rule

The Chain Rule enables us to differentiate a composite function f∘g. But why is there the need to have a special way to determine the derivative of a composite function? Intuitively, it is because the variation of the domain of f is now governed by the function g(x) rather than just by x, and g's rate of change with respect to x should somehow be taken into account. Before proceeding, see if you find effect of g by comparing the derivative of f(x)=x² with the derivative of f(x)=(5x)² where g(x)=5x.

---

The Chain Rule

We want to derive a rule for the derivative of a composite function of the form f∘g in terms of the derivatives of f and g. This rule would allow us to differentiate complicated functions in terms of known derivatives of simpler functions.

The rule that enables this is called the Chain Rule:

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

\[ (f∘g)′(x)=f′(g(x))g′(x) \nonumber\]

Or an equivalent statement:

If u=u(x) and f=f(u), then \[ \frac{d}{dx}[f(u)]=f′(u)\frac{du}{dx} \nonumber\]

Or another equivalent statement:

If y is a function of u, and u is a function of x, then

\[ \frac{dy}{dx}=\frac{dy}{du}⋅\frac{du}{dx} \nonumber\]

Apply the chain rule to find the derivative of \[ f(x)=(2x^3−4x^2+5)^2\nonumber\]

Using the chain rule, let \[ u=2x^3−4x^2+5.\nonumber\] Then

\[ \frac{d}{dx}[(2x^2−4x^2+5)^2]=\frac{d}{dx}[u^2] \nonumber\]

\[ =2u\frac{du}{dx} \nonumber\]

\[ =2(2x^3−4x^2+5)(6x^2−8x) \nonumber\]

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

\[ y=[u(x)]^n \nonumber\]

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

If \[ y=[u(x)]^n \nonumber\], then \[\frac{dy}{dx}=n[u(x)]^{n−1}\frac{d}{dx}u(x) \nonumber\]

---

Examples

Example 1

Earlier, you were asked if you find the effect of g on the derivative by comparing the derivative of f(x)=x² with the derivative of f(x)=(5x)² where g(x)=5x. The derivative of f(x)=x² is f′(x)=2x, and the derivative of f(x)=(5x)² is f′(x)=2(5x)5=2x⋅25. The effect of g in the composite function is to modify the rate of change of f(x)=x².

Example 2

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

We can write \[ y=(x2−3x+2)^{\frac{1}{2}} \nonumber\] This example illustrates the point that n can be any real number including fractions. Using the General Power Rule,

\[ \frac{dy}{dx}=\frac{1}{2}(x^2−3x+2)^{\frac{1}{2}-1}(2x−3) \nonumber\]

\[ =\frac{1}{2}(x^2−3x+2)^{−\frac{1}{2}}(2x−3) \nonumber\]

\[ =\frac{(2x−3)}{2\sqrt{x^2−3x+2}} \nonumber\]

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

\[ \frac{dy}{dx}|_{x=3}=\frac{2(3)−3}{2\sqrt{3^2−3(3)+2}}=\frac{3}{2\sqrt{2}}=\frac{3\sqrt{2}}{4} \nonumber\]

Example 3

Find \[\frac{dy}{dx} \nonumber\] for \[ y=sin^3x \nonumber\]

The function can be written as y=[sinx]3. Thus

\[ \frac{dy}{dx}=\frac{d}{dx}[sinx]^3 \nonumber\]

\[ =3[sinx]^2[cosx] \nonumber\]

\[ =3sinx^2cosx \nonumber\]

Example 4

Find \[\frac{dy}{dx} \nonumber\] for \[ y=[cos(πx^2)]^3 \nonumber\]

This example will show the application of the chain rule multiple times because there are several functions embedded within each other.

The function y can be written in the form

\[ y=(u(w))^3 \nonumber\] where

\[u(w)=cos(w) \nonumber\]

\[w(x)=πx^2 \nonumber\]

Here are the steps for the solution:

\[ \frac{dy}{dx}=\frac{d}{dx}[u(w)^3] \nonumber\]

…Use u and w substitutions

\[ =3⋅u(w)^2⋅\frac{du}{dx} \nonumber\]

.…After using the General Power Rule

\[ =3⋅u(w)^2⋅(\frac{du}{dw}⋅\frac{dw}{dx}) \nonumber\]

…After using the Chain Rule for ^du/_dx

\[ =3⋅u(w)^2⋅[−sin(w)⋅2πx] \nonumber\]

…After evaluating dudw and dwdx

\[ =3[cos(πx^2)]^2⋅(−sin(πx^2)⋅2πx) \nonumber\]

…After substituting for u and w

\[ =−6πx[cos(πx^2)]^2sin(πx^2) \nonumber\]

.…After simplification.

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

---

Review

For #1-11, find f′(x).

1. \[ f(x)=(2x^2−3x)^{39} \nonumber\]
2. \[ f(x)=(x^3−\frac{5}{x^2})^{−3} \nonumber\]
3. \[ f(x)=\frac{1}{\sqrt{3x^2−6x+2}} \nonumber\]
4. \[ f(x)=sin^3x \nonumber\]
5. \[ f(x)=sinx^3 \nonumber\]
6. \[ f(x)=sin^3x^3 \nonumber\]
7. \[ f(x)=tan(4x^5) \nonumber\]
8. \[ f(x)=\frac{4x−sin^2}{2x} \nonumber\]
9. \[ f(x)=\frac{sinx}{cos(3x−2)} \nonumber\]
10. \[ f(x)=(5x+8)^3(x^3+7x)^{13} \nonumber\]
11. \[ f(x)=(\frac{x−3}{2x−5})^3 \nonumber\]
12. Find \[ \frac{dy}{dx} \nonumber\] for \[ y=5cos(3x^2−1) \nonumber\]
13. Find the derivative of \[ \sqrt{x^3+x^5+89} \nonumber\]
14. Find the derivative of sin(sin(sin(x))).
15. By definition, any function composed with its inverse is just the identity: f(f⁻¹(x))=x. differentiate both sides of this equation and solve algebraically for the derivative of the inverse.

---

Review (Answers)

To see the Review answers, open this PDF file and look for section 3.8.

---

Vocabulary

Term	Definition
chain rule	The chain rule is the method for computing the derivative of a composite function. It states that for functions f(x) and g(x), (f∘g)′(x)=f′(g(x))g′(x).
composite function	A composite function is a function h(x) formed by using the output of one function g(x) as the input of another function f(x). Composite functions are written in the form h(x)=f(g(x)) or h=f∘g.
derivative	The derivative of a function is the slope of the line tangent to the function at a given point on the graph. Notations for derivative include f′(x), ^dy/dx, y′, ^df/dx and \frac{df(x)}{dx}.

---

Additional Resources

PLIX - Play, Learn, Interact, eXplore - Differentiation: Chain Rule

Video: Chain Rule Introduction

Practice: Chain Rule

Real World: Goodbye Waves