Inverse Function Theorem

The Derivative of an Inverse Function. $(f^{-1}{)}^{\prime }(x)=\frac{1}{f^{\prime }(f^{-1}(x))}$ $(f^{-1}{)}^{\prime \prime }(x)=\frac{f^{\prime \prime }(f^{-1}(x))}{(f^{\prime }(f^{-1}(x)){)}^3}$

Rigorous Theorem

1. Let $f$ be invertible
2. Let $f$ be differentiable
3. Let $f^{\prime }(x)\ne 0$ for all $x\in I$ Theorem conclusions:
4. Then, $f^{-1}$ is differentiable
5. Then $(f^{-1}{)}^{\prime }(x)=\frac{1}{f^{\prime }(f^{-1}(x))}$

Proving 4 (That $f^{-1}$ is differentiable)

Too hard at the moment

Proving 5 (That $(f^{-1}{)}^{\prime }(x)=\frac{1}{f^{\prime }(f^{-1}(x))}$)

1. Assume $f,f^{-1}$ are inverses of eachother
2. Assume $f$ is differentiable
3. Assume $f^{-1}$ is differentiable (but this should be proven)
4. And $f^{\prime }(x)\ne 0$ for all $x\in I$
5. Then $f(f^{-}1(x))=x$ From Inverse Function Cancellation
6. Then $f^{\prime }(f^{-1}(x))\ast (f^{-1})(x)=1$ From 5, Chain Rule
7. Then $f^{\prime }(x)=\frac{1}{f^{\prime }(f^{-1}(x))}$ Since $f^{\prime }(x)\ne 0$