Question

$f(x) = {x}^{2} + 4x - 5$ ; >-2
Find $\frac{d {f}^{ - 1} }{dx}$ at x=16​

Please show solving

Answer 1

The inverse function theorem says

$\dfrac{\mathrm df^{-1}}{\mathrm dx}(16)=\dfrac1{\frac{\mathrm df}{\mathrm dx}(f^{-1}(16))}$

We have

$f(x)=x^2+4x-5$

defined on $x>-2$, for which we get

$f^{-1}(x)=-2+\sqrt{x+9}$

and

$f^{-1}(16)=-2+\sqrt{16+9}=3$

The derivative of $f(x)$ is

$f'(x)=2x+4$

So we end up with

$\dfrac{\mathrm df^{-1}}{\mathrm dx}(16)=\dfrac1{\frac{\mathrm df}{\mathrm dx}(3)}=\dfrac1{10}$