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. Twice differentiating the identity with respect to x obtains,