## Calculus AB and Calculus BC

## CHAPTER 3 Differentiation

### C. THE CHAIN RULE; THE DERIVATIVE OF A COMPOSITE FUNCTION

Formula (3) says that

This formula is an application of the *Chain Rule.* For example, if we use formula (3) to find the derivative of (*x*^{2} − *x* + 2)^{4}, we get

In this last equation, if we let *y* = (*x*^{2} − *x* + 2)^{4} and let *u* = *x*^{2} − *x* + 2, then *y* = *u*^{4}. The preceding derivative now suggests one form of the Chain Rule:

as before. Formula (3) the previous page gives the general case where *y* = *u** ^{n}* and

*u*is a differentiable function of

*x*.

Now suppose we think of *y* as the composite function *f* (*g*(*x*)), where *y* = *f* (*u*) and *u* = *g*(*x*) are differentiable functions. Then

**Chain rule**

as we obtained above. The Chain Rule tells us how to differentiate the composite function: “Find the derivative of the ‘outside’ function first, then multiply by the derivative of the ‘inside’ one.”

For example:

Many of the formulas listed above in §B and most of the illustrative examples that follow use the Chain Rule. Often the chain rule is used more than once in finding a derivative.

Note that the algebraic simplifications that follow are included only for completeness.

**EXAMPLE 1**

If *y* = 4*x*^{3} − 5*x* + 7, find *y* *′*(1) and *y* *″*(1).

**SOLUTION:**

Then *y* *′*(1) = 12 · 1^{2} − 5 = 7 and *y* *″*(1) = 24 · 1 = 24.

**EXAMPLE 2**

If *f* (*x*) = (3*x* + 2)^{5}, find *f* *′*(*x*).

**SOLUTION:** *f* *′*(*x*) = 5(3*x* + 2)^{4} · 3 = 15(3*x* + 2)^{4}.

**EXAMPLE 3**

**SOLUTION:**

**EXAMPLE 4**

**SOLUTION:**

**EXAMPLE 5**

If *s*(*t*) = (*t*^{2} + 1)(1 − *t*)^{2}, find *s* *′*(*t*).

**SOLUTION:**

**EXAMPLE 6**

If *f* (*t*) = *e*^{2}* ^{t}* sin 3

*t*, find

*f*

*′*(0).

**SOLUTION:**

Then, *f* *′*(0) = 1(3 · 1 + 2 · 0) = 3.

**EXAMPLE 7**

**SOLUTION:**

Note that neither *f* (*v*) nor *f* *′*(*v*) exists where the denominator equals zero, namely, where 1 − 2*v*^{2} = 0 or where *v* equals

**EXAMPLE 8**

If *x* ≠ 0, find *f* *′*(*x*).

**SOLUTION:**

**EXAMPLE 9**

If *y* = tan (2*x*^{2} + 1), find *y* *′*.

**SOLUTION:** *y* *′* = 4*x* sec^{2} (2*x*^{2} + 1).

**EXAMPLE 10**

If *x* = cos^{3} (1 − 3θ), find

**SOLUTION:**

**EXAMPLE 11**

If *y* = *e*^{(sin} ^{x}^{) + 1}, find

**SOLUTION:**

**EXAMPLE 12**

If *y* = (*x* + 1)ln^{2}(*x* + 1), find

**SOLUTION:**

**EXAMPLE 13**

If *g*(*x*) = (1 + sin^{2} 3*x*)^{4}, find

**SOLUTION:**

Then

**EXAMPLE 14**

If *y* = sin^{−1} *x* + find *y* *′*.

**SOLUTION:**

**EXAMPLE 15**

If *u* = ln find

**SOLUTION:** .

**EXAMPLE 16**

If *s* = *e** ^{−t}*(sin

*t*− cos

*t*), find

*s*

*′*.

**SOLUTION:**

**EXAMPLE 17**

Let *y* = 2*u*^{3} − 4*u*^{2} + 5*u* − 3 and *u* = *x*^{2} − *x*. Find

**SOLUTION:**

**EXAMPLE 18**

If *y* = sin (*ax* + *b*), with *a* and *b* constants, find

**SOLUTION:** = [cos(*ax* + *b*)] · *a* = *a* cos(*ax* + *b*).

**EXAMPLE 19**

If *f* (*x*) = *ae** ^{kx}* (with

*a*and

*k*constants), find

*f*

*′*and

*f*

*″*.

**SOLUTION:** *f* *′*(*x*) = *kae** ^{kx}* and

*f*

*″*=

*k*

^{2}

*ae*

*.*

^{kx}**EXAMPLE 20**

If *y* = ln (*kx*), where *k* is a constant, find

**SOLUTION:** We can use both formula (13), and the Chain Rule to get

Alternatively, we can rewrite the given function using a property of logarithms: ln (*kx*) = ln *k* + ln *x*. Then

**EXAMPLE 21**

Given *f* (*u*) = *u*^{2} − *u* and *u* = *g*(*x*) = *x*^{3} − 5 and *F*(*x*) = *f* (*g*(*x*)), evaluate *F* *′*(2).

**SOLUTION:** *F* *′*(2) = *f* *′*(*g*(2))*g* *′*(2) = *f* *′*(3) · (12) = 5 · 12 = 60.

Now, since *g* *′*(*x*) = 3*x*^{2}, *g* *′*(2) = 12, and since *f* *′*(*u*) = 2*u* − 1, *f* *′*(3) = 5. Of course, we get exactly the same answer as follows.