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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