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Derivative of the inverse function f^-1 is given by : (f^-1)’(x)=1 / f’ (f^-1(x))

To prove this result, we are going to apply the Chain rule (derivative of a composite function) to the function f and to its inverse f^-1.

We previously demonstrated the result relating to the Chain Rule.

Consider $I$ and $J$ two intervals of $\mathbb {R} $ and two functions $u, v $ defined by

$$ \begin{aligned} u&: I \rightarrow \mathbb{R}\\ v&: J \rightarrow \mathbb{R} \end{aligned} $$

such $f(I) \subset J$. Let $x$ a point of the interval $I$.

If $u$ is differentiable at $x$ and $v$ is differentiable at $u(x)$ then the composite function $u \circ v$ is differentiable at $x$, and the Chaine Rule is given by

$$ \forall x\in I, \quad \left(u \circ v\right)^{\prime}(x)=u^{\prime}(v(x)) \cdot v^{\prime}(x) $$

By applying the Chain Rule derivative of composite function avec $u=f,v=f^{-1}$, on a alors:

$$ \left(f \circ f^{-1}\right)^{\prime}(x)=f^{\prime}(f^{-1}(x)) \cdot (f^{-1})^{\prime}(x) $$

However, by definition of a reciprocal function:

$$ \left(f \circ f^{-1}\right)(x)=\left(f^{-1} \circ f\right)(x)=Id(x)=x $$

where $Id$ is the identity function.

Then:

$$ \left(f \circ f^{-1}\right)^{\prime}(x)=\left(Id\right)^{\prime}(x)=1 $$

Thus

$$ \begin{aligned} \left(f \circ f^{-1}\right)^{\prime}(x)&=f^{\prime}(f^{-1}(x)) \cdot (f^{-1})^{\prime}(x)\\ &=1 \end{aligned} $$

We conclude:

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