Inverse functions and differentiation

Rule:

{\color {CornflowerBlue}{f'}}(x)={\frac {1}{{\color {Salmon}{(f^{{-1}})'}}({\color {Blue}{f}}(x))}}

Example for arbitrary

x_{0}\approx 5.8

{\color {CornflowerBlue}{f'}}(x_{0})={\frac {1}{4}}

{\color {Salmon}{(f^{{-1}})'}}({\color {Blue}{f}}(x_{0}))=4~

In mathematics, the inverse of a function $y=f(x)$ is a function that, in some fashion, "undoes" the effect of $f$ (see inverse function for a formal and detailed definition). The inverse of $f$ is denoted $f^{-1}$ . The statements y = f(x) and x = f⁻¹(y) are equivalent.

Their two derivatives, assuming they exist, are reciprocal, as the Leibniz notation suggests; that is:

{\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}=1.

This is a direct consequence of the chain rule, since

{\frac {dx}{dy}}\,\cdot \,{\frac {dy}{dx}}={\frac {dx}{dx}}

and the derivative of $x$ with respect to $x$ is 1.

Writing explicitly the dependence of $y$ on $x$ and the point at which the differentiation takes place and using Lagrange's notation, the formula for the derivative of the inverse becomes

\left[f^{{-1}}\right]'(a)={\frac {1}{f'\left(f^{{-1}}(a)\right)}}

Geometrically, a function and inverse function have graphs that are reflections, in the line y = x. This reflection operation turns the gradient of any line into its reciprocal.

Assuming that $f$ has an inverse in a neighbourhood of $x$ and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at $x$ and have a derivative given by the above formula.

Examples

$\,y=x^{2}$ (for positive $x$ ) has inverse $x={\sqrt {y}}$ .

{\frac {dy}{dx}}=2x{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{2{\sqrt {y}}}}={\frac {1}{2x}}

{\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=2x\cdot {\frac {1}{2x}}=1.

At x = 0, however, there is a problem: the graph of the square root function becomes vertical, corresponding to a horizontal tangent for the square function.

$\,y=e^{x}$ (for real $x$ ) has inverse $\,x=\ln {y}$ (for positive $y$ )

{\frac {dy}{dx}}=e^{x}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{y}}

{\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=e^{x}\cdot {\frac {1}{y}}={\frac {e^{x}}{e^{x}}}=1

Additional properties

Integrating this relationship gives

{f^{{-1}}}(x)=\int {\frac {1}{f'({f^{{-1}}}(x))}}\,{dx}+c.

This is only useful if the integral exists. In particular we need

f'(x)

to be non-zero across the range of integration.

It follows that a function that has a continuous derivative has an inverse in a neighbourhood of every point where the derivative is non-zero. This need not be true if the derivative is not continuous.

Higher derivatives

The chain rule given above is obtained by differentiating the identity x = f⁻¹(f(x)) with respect to x. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to x , one obtains