# Derivative of inverse of f(x)=e^(-2x)-9x^3+4 at point (0,5)?

Jul 7, 2018

$- \frac{1}{2}$

#### Explanation:

Derivative of $f \left(x\right)$ :

${f}^{'} \left(x\right) = \frac{d}{\mathrm{dx}} \left({e}^{- 2 x} - 9 {x}^{3} + 4\right) = - 2 {e}^{- 2 x} - 27 {x}^{2}$

Thus

${f}^{'} \left(0\right) = - 2$

Let $g$ be the inverse of $f$. Then

$g \left(f \left(x\right)\right) = x$

Using the chain rule, we get

${g}^{'} \left(f \left(x\right)\right) \times {f}^{'} \left(x\right) = 1 \implies$

${g}^{'} \left(f \left(0\right)\right) \times \left(- 2\right) = 1 \implies$

${g}^{'} \left(5\right) = - \frac{1}{2}$