Inverse Derivative/Proof

To prove that $f^{-1}$ is differentiable given that $f$ is invertible and differentiable, we will use the definition of derivative. Let $y_0 \in \R$, so that $f(x_0)=y_0$ for some $x_0 \in \R$. Since $f$ is differentiable it is continuous, so if $x \to x_0$ we have $y=f(x)\to f(x_{0})=y_{0}$. Now, using the definition of derivative:

${\begin{aligned}(f^{-1})'(y_{0})&=\lim _{y\to y_{0}}{\frac {f^{-1}(y)-f^{-1}(y_{0})}{y-y_{0}}}\\&=\lim _{y\to y_{0}}{\frac {f^{-1}(f(x))-f^{-1}(f(x_{0}))}{f(x)-f(x_{0})}}\\&=\lim _{x\to x_{0}}{\frac {x-x_{0}}{f(x)-f(x_{0})}}\\&={\frac {1}{f'(x_{0})}}\\&={\frac {1}{f'(f^{-1}(x_{0}))}}.\end{aligned}}$