# Let f(x) = x^2 - 4x - 5, x > 2, how do you find the value of (df^-1)/dx (or the derivative of the inverse of f(x)), at the point x = 0 = f(5)?

Aug 8, 2016

$= \frac{1}{6}$

#### Explanation:

we have $f \left(x\right) = {x}^{2} - 4 x - 5$ and you want the derivative of its inverse

so if

$y = f \left(x\right) , x = g \left(y\right)$, you are actually looking for $\frac{d}{\mathrm{dy}} g \left(y\right) = g ' \left(y\right)$ is the derivative of the inverse with respect to its independent variable.

to that end we can say that

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

and using the chain rule:

$\frac{d}{\mathrm{dx}} \left(x = g \left(f \left(x\right)\right)\right)$

$\implies 1 = g ' \left(y\right) f ' \left(x\right)$

so $g ' \left(y\right) = {\left({f}^{- 1} \left(x\right)\right)}^{p} r i m e = = \frac{1}{f ' \left(x\right)}$
$= \frac{1}{2 x - 4}$

$= \frac{1}{6}$

there's poss another way to show this, which is a bit of a desecration of the Liebnitz notation: ie $\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}$, really not sure that means very much though as $\frac{\mathrm{dy}}{\mathrm{dx}}$ is not a fraction with a numerator and denominator