Answer
$\frac{1}{3}$
Work Step by Step
Given: $f(x)=x^3+3\sin x+2\cos x$
To find $(f^{-1})'(2)$, we need to find $f^{-1}(2)$ and $f'(f^{-1}(2))$.
Find $f^{-1}(2)$:
$f(0)=0^3+3\sin 0+2\cos 0$
$f(0)=0+3\cdot 0+2\cdot 1$
$f(0)=2$
$f^{-1}(2)=0$
Find $f'(0)$:
$f'(x)=\frac{d}{dx}(x^3+3\sin x+2\cos x)$
$f'(x)=3x^2+3\cos x+2(-\sin x)$
$f'(x)=3x^2+3\cos x-2\sin x$
$f'(0)=3\cdot 0^2+3\cos 0-2\sin 0$
$f'(0)=3\cdot 0+3\cdot 1-2\cdot 0$
$f'(0)=3$
Using the Formula in Exercise 83,
$(f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(0)}=\frac{1}{3}$
