Question

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)?

Answer 1

`=1/6`

Explanation:

we have `f(x) = x^2 - 4x - 5` and you want the derivative of its inverse

so if

`y = f(x), x = g(y)`, you are actually looking for `d/dy g(y) = g'(y)` is the derivative of the inverse with respect to its independent variable.

to that end we can say that

`x = g (f(x))`

and using the chain rule:

`d/dx ( x = g (f(x)) )`

`implies 1 = g' (y) f'(x)`

so `g'(y) =(f^(-1)(x))^prime = = 1/(f'(x))`
`=1/( 2x - 4)`

`=1/6`

there's poss another way to show this, which is a bit of a desecration of the Liebnitz notation: ie `dx/dy = 1/ (dy/dx)`, really not sure that means very much though as `dy/dx` is not a fraction with a numerator and denominator