Derivative of the inverse

Example

Consider the function

$f(x)=3x+2e^x.$

$f$ is defined and differentiable at every $\mathbb{R}$, and is strictly increasing (because it is the sum of two strictly increasing functions).
So it has an inverse function, which we indicate with the usual symbol $f^{-1}$, also defined on the whole $\mathbb{R}$.
Suppose we want to know the value of the derivative of $f^{-1}$ at $x=2$. The formula for the derivative of the inverse is this:

$[f^{-1}]'(2)=\frac{1}{f'(a)},$

where $a$ is the only value that satisfies 

$a=f^{-1}(2)\quad\text{or equivalently}\quad f(a)=2.$

It is easy to observe that $f(0)=2$, so we deduce that $a=0$. So we will have

$(*)\qquad[f^{-1}]'(2)=\frac{1}{f'(0)}.$

We now need the derivative of $f$, calculated at 0: using the linearity property of the derivative we have

$f'(x)=3\frac{d}{dx}x+2\frac{d}{dx}e^x=3+2e^x,$

so for $x=0$ we find

$f'(0)=3+2e^0=3+2=5.$

Now just substitute in the formula $(*)$ to get the result:

$\qquad[f^{-1}]'(2)=\frac{1}{5}.$