# Thread: Inverse functions + derivative

1. ## Inverse functions + derivative

Hi, I have a question that is troubling me:

Given that for f: R -> R such that f(x) = 2 x + cos x, find g' (1) where g is the inverse of f.

I've tried trying to find g but I get stuck at: x = 2y + cos y.

If anyone could help, it would be greatly appreciated.

Cheers.

2. Originally Posted by shinn
Hi, I have a question that is troubling me:

Given that for f: R -> R such that f(x) = 2 x + cos x, find g' (1) where g is the inverse of f.

I've tried trying to find g but I get stuck at: x = 2y + cos y.

If anyone could help, it would be greatly appreciated.

Cheers.
You're on a hiding to nothing trying to first get the rule for the inverse.

Clearly the inverse is defined as an implicit function of $\displaystyle x$ by $\displaystyle x = 2y + \cos y$. Therefore, using implicit differentiation:

$\displaystyle 1 = 2 \frac{dy}{dx} - \sin y \frac{dy}{dx}$.

So $\displaystyle 1 = 2 g'(x) - g'(x) \sin g(x)$. You require the value of $\displaystyle g'(1)$: $\displaystyle 1 = 2 g'(1) - g'(1) \sin g(1)$.

Note that the value of g(1) is given by the solution to $\displaystyle 1 = 2y + \cos y$. By inspection, $\displaystyle y = 0$, that is, $\displaystyle g(1) = 0$.

Therefore ....