304
Chapter 3 | Derivatives
Now let g ( x ) =2 x 3 , so g ′( x ) =6 x 2 . Substituting into the previous result, we obtain h ′( x ) = 1 1−4 x 6 ·6 x 2 = 6 x 2 1−4 x 6 .
Use the inverse function theorem to find the derivative of g ( x ) = tan −1 x .
3.45
The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem. These formulas are provided in the following theorem.
Theorem 3.13: Derivatives of Inverse Trigonometric Functions
d dx sin d dx cos
(3.22)
−1 x = 1
1−( x ) 2
(3.23)
−1 x = −1
1−( x ) 2
d dx tan d dx cot
(3.24)
−1 x = 1
1+( x ) 2
(3.25)
−1 x = −1
1+( x ) 2
d dx sec
(3.26)
−1 x = 1
| x | ( x ) 2 −1
d dx csc
(3.27)
−1 x = −1
| x | ( x ) 2 −1
Example 3.65 Applying Differentiation Formulas to an Inverse Tangent Function
Find the derivative of f ( x ) = tan −1 ⎛ ⎝ x 2 ⎞ ⎠ .
Solution Let g ( x ) = x 2 , so g ′( x ) =2 x . Substituting into Equation 3.24 , we obtain f ′( x ) = 1 1+ ⎛ ⎝ x 2 ⎞ ⎠ 2 · (2 x ).
Simplifying, we have
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