Inversion Formula
How to calculate the derivative of the inverse function.
Discover Bridges
Study Inversion Formula
Theorem 1 (Differentiation of inverse functions). Let \(I,J\subset\mathbb{R}\) be intervals and let \(f:I\to J\) be bijective. If \(f\) is differentiable in \(x_0\in I\) with \(f'(x_0)\neq0\) and if \(f^{-1}:J\to I\) is continuous in \(y_0=f(x_0)\), then the inverse function \(f^{-1}\) is differentiable in \(y_0:=f(x_0)\) with \[\left(f^{-1}\right)'(y_0)=\frac{1}{f'(f^{-1}(y_0))}.\]
Example 2.
\(g(x)=\sqrt[n]{x}=x^{\frac{1}{n}}\) is the inverse function of \(f:\mathbb{R}^+\to\mathbb{R}^+\), \(f(x)=x^n\). Then for \(x>0\) holds \[g'(x)=\frac{1}{f'(g(x))}=\frac{1}{n(g(x))^{n-1}}=\frac{1}{n\sqrt[n]{x}^{n-1}}=\frac1nx^{-\frac{n-1}{n}}=\frac1nx^{\frac{1}{n}-1}.\]
\(\log: (0,\infty)\to\mathbb{R}\) is the inverse function of \(\exp\). Then for \(x>0\) holds \[\log'(x)=\frac1{\exp'(\log(x))}=\frac1{\exp(\log(x))}=\frac1{x}.\]
\(\arcsin:[-1,1]\to[-\frac\pi2,\frac\pi2]\) is the inverse function of \(\sin\). Then for \(x\in (-\frac\pi2,\frac\pi2)\) holds \[\arcsin'(x)=\frac1{\sin'(\arcsin(x))}=\frac1{\cos(\arcsin(x))}.\] Since \(\sin^2(y)+\cos^2(y)=1\) and \(\cos(y)\geq0\) for \(y\in (-\frac\pi2,\frac\pi2)\), we have \[\cos(y)=\sqrt{1-\sin^2(y)}.\] Therefore \[\arcsin'(x)=\frac1{\sqrt{1-\sin^2(\arcsin(x))}}=\frac1{\sqrt{1-x^2}}.\]
Discuss your questions by typing below.
Solve the WeBWorK Exercise
The data for the interactive network on this webpage was generated with pntfx Copyright Fabian Gabel and Julian Großmann. pntfx is licensed under the MIT license. Visualization of the network uses the open-source graph theory library Cytoscape.js licensed under the MIT license.