User:TigerTjäder/Math2

Suppose $$f^{-1}$$ is the inverse function of a differentiable function $$f$$ and let $$G(x) = 1/f^{-1}(x)\!$$. If $$f(3) = 2$$ and $$f'(3) = 1/9$$, find $$G'(2)$$

We are gonna use two identities:

(1) $$(f^{-1})'(a) = \frac{1}{f'(f^{-1}(a))}$$

and

(2) $$f(x) = y \iff f^{-1}(y) = x$$


 * $$G'(2)$$
 * $$G'(x) = \frac{-(f^{-1})'(x)}{f^{-1}(x)^2} (f^{-1})'(x) = \frac{-(f^{-1})'(x)^2}{(f^{-1}(x)^2)}$$ (Chain rule)
 * $$G'(2) = \frac{-(f^{-1})'(2)^2}{f^{-1}(2)^2}$$
 * $$f(3) = 2 \iff f^{-1}(2) = 3$$ (Identity (1))
 * $$G'(2) = \frac{-(f^{-1})'(2)^2}{3^2}$$
 * $$G'(2) = \frac{-(f^{-1})'(2)^2}{9}$$
 * $$(f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))}$$ (Identity (2), with $$a = 2$$)
 * $$f(3) = 2 \iff f^{-1}(2) = 3$$ (Identity (1))
 * $$(f^{-1})'(2) = \frac{1}{f'(3)}$$
 * $$f'(3) = 1/9$$
 * $$(f^{-1})'(2) = \frac{1}{\frac{1}{9}}$$
 * $$(f^{-1})'(2) = 9$$
 * $$G'(2) = \frac{-9^2}{9}$$
 * $$G'(2) = \frac{-81}{9}$$
 * $$G'(2) = -9$$