MATH 136 More Derivatives: Product Rule, Quotient Rule, Chain Rule

The Product Rule gives the formula for differentiating the product of two functions, and the Quotient Rule gives the formula for differentiating the quotient of two functions.

(uv ′ ) = ′ u v + u ′ v

Product Rule

uv

' =

′ u v − u ′ v

v2

Quotient Rule Example 1. Find the derivative of each function and simplify:

(a) f ( x) = 4x3 sin x (b) g(x ) = e3x cos x (c) h(x ) = x ln x

Solution. For each, we use the product rule. (a) ′ f (x ) = 12 x2 sin x + 4 x3 cos x (b) ′ g ( x) = 3e3x cos x + e3x (− sin x) = e3x (3cos x − sin x )

(c) ′ h ( x) =xxln x + x

1x

=x ln x +1( )

x

Example 2. Find the derivative of each function and simplify:

(a) f ( x) = 8sin x3x4 (b) g(x ) = cos(4x )

x (c) h(x ) = 4 x

2

e4x

Solution. For each, we use the quotient rule:

(a) ′ f (x ) =

(8 cosx) × 3x4 − (8sin x) ×12x3

3x4( )2 = 24 x 4 cos x − 96 x3 sin x

9x8 = 8 x cos x − 32sin x

3x5

(b) ′ g ( x) =

−4sin(4x ) × x − cos(4x ) × (1 / 2)x−1/2

( x )2=−4 x sin(4x ) − cos(4 x )

2 xx

=−8x sin(4x ) − cos(4x )

2x3/2

(c) ′ h ( x) =8x × e4x − 4 x2(4e4x )

(e4x )2=

e4x 8x −16 x2( )(e4x )2

=8x − 16x 2

e4x

The Chain Rule The Chain Rule gives the process for differentiating a composition of functions:

If y = f (g(x)) , then

dydx

= ′ f (g(x)) × ′ g (x) .

We have already used the Chain Rule for functions of the form y = f (mx) to obtain ′ y = m × ′ f (mx) . For instance, if y = sin(4 x) , then ′ y = 4 cos(4x ) . But the Chain Rule

applies to a general composition of functions y = f (g(x)) , where g(x ) is not just the linear function m x .

Example 3. Let f (x) = 9x3 − 8sec x( )4 . Find ′ f (x) .

Solution: The function is of the form y = g( x)( )4 . So ′ y = 4 g(x )( )3 × ′ g ( x) . Thus we

have ′ f (x) = 4 9x3 − 8sec x( )3 × 27 x2 − 8sec x tan x( ). Example 4. Let f (x) =

16

25 − 4x2( )3. Find ′ f (x) .

Solution. We first re-write f as f (x) = 16 25 − 4x2( )−3. Then by the Chain Rule,

′ f (x) = −48 25 − 4x2( )−4× (−8x) =

384 x

25 − 4x2( )4.

Example 5. Let f (x) = sin

3(2 πx) . Find ′ f (x) . Solution. We first re-write f as f (x) = sin(2π x)( )3 . Then applying the Chain Rule twice, we obtain

′ f (x) = 3 sin(2π x)( )2 ×cos(2 π x)× (2π) = 6π sin2(2 πx) cos(2 πx ) .

From Example 5, we see that we may have to apply the Chain Rule more than once when we have a function of the form y = f (g(h(x ))) . Here, we have

′ y = ′ f (g(h(x ))) × ′ g (h(x )) × ′ h (x ) ,

which is “the derivative of the outside times the derivative of the inside times the derivative of the next inside . . . ” Example 6. Let f (x) = cos(4x) . Find ′ f (x) . Solution. Applying the Chain Rule twice, we obtain

′ f (x) = cos(4x)cos(4x)

× (−sin(4x))× 4 = −4 cos(4x) tan(4x).

Example 7. Let f (x) = 4

sec2 x + tan2 x. Find ′ f (x).

Solution. We first re-write f as f (x) = 4 sec2(x) + tan2(x)( )−1/ 2

. Then applying the Chain Rule, we first have

′ f (x) = −2 sec2 x + tan2 x( )−3/2 × ddx

sec2 x + tan2 x( ) ;

but we still must take the derivative of the “inner function” sec2 x + tan2 x . We apply the Chain Rule to each term: 2sec x × (sec x tan x) + 2 tan x × (sec2 x ); thus,

′ f (x) = −2

sec2 x + tan2 x( )3/2× 2sec2 x tan x + 2 tan x sec2 x( ) = −8 sec2 x tan x

sec2 x + tan2 x( )3/2.

Some derivatives require using a combination of the Product, Quotient, and Chain rules.

Example 8. Let

f (x) = sin(4x 3)

9 − 3x2( )6. Find ′ f (x) .

Solution. We apply the Quotient Rule, but use the Chain Rule when differentiating the numerator and the denominator:

′ f (x) =

cos(4x3) × (12x2 ) × 9 − 3x 2( )6 − sin(4x3) × 6 9 − 3x2( )5 × (−6x)

9 − 3x2( )12

=

12x2(9 − 3x2) cos(4x3 ) + 36x sin(4x 3)

9 − 3x2( )7. (After factoring out (9 − 3x2)5 .)

Example 9. Let f (x) = tan

4(2x) 9 − x2 . Find ′ f (x) . Solution. We first re-write f as f (x) = (tan(2x))

4 × (9 − x2)1/ 2. Then ′ f (x) =

4(tan(2x))3 × sec2(2x) × 2 × (9 − x2)1/ 2 + (tan(2x))4 × 1

2(9 − x2 )−1/ 2 × (−2x )

= 8 tan3(2x) sec2 (2x) 9 − x 2 −

x tan4(2x)9 − x2

=8 tan3(2x ) sec2 (2x ) 9 − x2( ) − x tan4 (2x )

9 − x2 .