Calculus Volume 1

Chapter 3 | Derivatives

291

Think of h ( x ) =cos( g ( x )) as f ⎛

⎞ ⎠

⎝ g ( x )

f ( x ) =cos x . Since f ′( x ) =−sin x . we have

where

f ′ ⎛

⎞ ⎠ =−sin ⎛

⎞ ⎠ . Then we do the following calculation. h ′( x ) = f ′ ⎛ ⎝ g ( x ) ⎞ ⎠ g ′( x )

⎝ g ( x )

⎝ g ( x )

Apply the chain rule.

=−sin ⎛

⎞ ⎠ g ′( x ) Substitute f ′ ⎛

⎞ ⎠ =−sin ⎛

⎞ ⎠ .

⎝ g ( x )

⎝ g ( x )

⎝ g ( x )

Thus, the derivative of h ( x ) =cos ⎛

⎞ ⎠ is given by h ′( x ) =−sin ⎛

⎞ ⎠ g ′( x ).

⎝ g ( x )

⎝ g ( x )

In the following example we apply the rule that we have just derived.

Example 3.52 Using the Chain Rule on a Cosine Function

Find the derivative of h ( x ) =cos ⎛

⎞ ⎠ .

⎝ 5 x 2

Solution Let g ( x ) =5 x 2 . Then g ′( x ) =10 x . Using the result from the previous example, h ′( x ) =−sin ⎛ ⎝ 5 x 2 ⎞ ⎠ ·10 x =−10 x sin ⎛ ⎝ 5 x 2 ⎞ ⎠ .

Example 3.53 Using the Chain Rule on Another Trigonometric Function

Find the derivative of h ( x ) = sec ⎛

⎝ 4 x 5 +2 x ⎞ ⎠ .

Solution Apply the chain rule to h ( x ) = sec ⎛

⎞ ⎠ to obtain h ′( x ) = sec( g ( x )tan ⎛

⎝ g ( x )

⎝ g ( x ) ⎞ ⎠ g ′( x ). In this problem, g ( x ) =4 x 5 +2 x , so we have g ′( x ) =20 x 4 +2. Therefore, we obtain h ′( x ) = sec ⎛ ⎝ 4 x 5 +2 x ⎞ ⎠ tan ⎛ ⎝ 4 x 5 +2 x ⎞ ⎠ ⎛ ⎝ 20 x 4 +2 ⎞ ⎠

= (20 x 4 +2)sec ⎛

⎝ 4 x 5 +2 x ⎞

⎛ ⎝ 4 x 5 +2 x ⎞ ⎠ .

⎠ tan

Find the derivative of h ( x ) = sin(7 x +2).

3.36

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