> [!NOTE] Theorem (Derivative of Inverses) > Let $f:(a,b)\to \mathbb{R}$ be a [[Fréchet Differentiation|differentiable real function]] with positive derivative. Then its inverse is differentiable and satisfies $(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))}$. ###### Proof Follows from chain rule ###### Proof from limit definition By [[Positive Derivative Implies Strictly Increasing Real Function]], since $f$ has a positive derivative, its [[Positive Derivative Implies Strictly Increasing Real Function|strictly increasing]] and [[Differentiablity implies Continuity|continuous]]. Therefore by [[Inverse of Strictly Monotonic Continuous Real Function Exists, is Strictly Monotonic in the Same Sense, and is Continuous]] it has an inverse that is continuous and strictly increasing. Let $(c,d)$ be the range of $f$, $x\in(c,d)$ and $g(x)=y$. We want to calculate $g'(x)= \lim_{ u \to x } \frac{g(u)-g(x)}{u-x}.$Let $v=g(u)$ so that $u=f(v)$. Then the quotient is $\frac{v-y}{f(v)-f(y)}.$As $u\to x$ we have $v=g(u)\to y=g(x)$ since $g$ is continuous at $x$. So we want to calculate $\lim_{ v \to y } \frac{v-y}{f(v)-f(y)} $We know that $\lim_{ v \to y } \frac{f(v)-f(y)}{v-y}=f'(y)$and that this limit is positive. So by [[Algebra of Limits of Convergent Sequences|algebra of limits]], we have $g'(x) = \frac{1}{f'(g(x))} $