Chapter 3 | Derivatives
303
Analysis To see that cos ⎛
⎞ ⎠ = 1− x 2 , consider the following argument. Set sin −1 x = θ . In this case, sin θ = x
⎝ sin −1 x
where − π 2 ≤ Since θ is an acute angle, we may construct a right triangle having acute angle θ , a hypotenuse of length 1 and the side opposite angle θ having length x . From the Pythagorean theorem, the side adjacent to angle θ has length 1− x 2 . This triangle is shown in Figure 3.29 . Using the triangle, we see that cos ⎛ ⎝ sin −1 x ⎞ ⎠ =cos θ = 1− x 2 . θ ≤ π 2 . We begin by considering the case where 0< θ < π 2 .
Figure 3.29 Using a right triangle having acute angle θ , a hypotenuse of length 1, and the side opposite angle θ having length x , we can see that cos ⎛ ⎝ sin −1 x ⎞ ⎠ =cos θ = 1− x 2 .
In the case where − π
2 < θ <0, we make the observation that 0<− θ < π 2
and hence
⎛ ⎝ sin −1 x
⎞ ⎠ =cos θ =cos(− θ ) = 1− x 2 .
cos
Nowif θ = π 2
or θ = − π
2 , x =1 or x =−1, and since in either case cos θ =0 and 1− x 2 =0, wehave cos ⎛ ⎝ sin −1 x ⎞ ⎠ =cos θ = 1− x 2 .
- =1
Finally, if θ = - , x =0 and cos θ = 1
.
Consequently, in all cases, cos ⎛
⎞ ⎠ = 1− x 2 .
⎝ sin −1 x
Example 3.64 Applying the Chain Rule to the Inverse Sine Function
Apply the chain rule to the formula derived in Example 3.61 to find the derivative of h ( x ) = sin −1 ⎛ ⎝ g ( x ) ⎞ ⎠ and use this result to find the derivative of h ( x ) = sin −1 ⎛ ⎝ 2 x 3 ⎞ ⎠ .
Solution Applying the chain rule to h ( x ) = sin −1 ⎛
⎞ ⎠ , we have h ′( x ) = 1 1− ⎛
⎝ g ( x )
g ′( x ).
⎞ ⎠ 2
⎝ g ( x )
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