f(x)=x-tanx
I am supposed to find the derivative to the inverse function in $(\pi$/4)$-1$. So my plan was to use the formula for the derivative of the inverse function. I have defined the domain to f(x) as $(-\pi/2,\pi/2)$, to make it a one-to-one function. But I'm not able to solve this equation
x-tanx=$(\pi/4)-1$
Could someone help me?
Recall that if $g(x)$ is the inverse of $f(x)$, at $x=x_0$ we have:
$$g'(x_0)=\frac1{f'(g(x_0))}$$
with $g(x_0)=\pi/4$ indeed $f(x)$ and $g(x)$ are symmetric with respect to the line $y=x$ and thus
$$f(\pi/4)=\pi/4-1\implies g(f(\pi/4))=\pi/4=g(\pi/4-1)$$