# Inverse functions and differentiation

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

Differentiation in calculus is the process of obtaining a derivative. The derivative of a function gives the slope at any point.

[itex]\frac{dy}{dx} [itex] denotes the derivative of the function [itex]y=f(x)[itex] with respect to [itex]x[itex].

[itex]\frac{dx}{dy} [itex] denotes the derivative of the function [itex]x=f(y)[itex] with respect to [itex]y[itex].

The two derivatives are, as the Leibniz notation suggests, reciprocal, that is

[itex]\frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = 1 [itex]

This is a direct consequence of the chain rule, since

[itex] \frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = \frac{dx}{dx} [itex]

and the derivative of [itex] x [itex] with respect to [itex] x [itex] is 1. 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.

## Examples

• [itex]y = x^2[itex] (for positive [itex]x[itex]) has inverse [itex]x = \sqrt{y}[itex].
[itex] \frac{dy}{dx} = 2x

\mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{dx}{dy} = \frac{1}{2\sqrt{y}} [itex]

[itex] \frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = 2x . \frac{1}{2\sqrt{y}} = \frac{2x}{2x} = 1 [itex]

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.

• [itex]y = e^x[itex] has inverse [itex] x = \ln (y)[itex] (for positive [itex]y[itex]).
[itex] \frac{dy}{dx} = e^x

\mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{dx}{dy} = \frac{1}{y} [itex]

[itex] \frac{dy}{dx}\,.\,\frac{dx}{dy} = e^x . \frac{1}{y} = \frac{e^x}{e^x} = 1 [itex]

• Integrating this relationship gives
[itex]{f^{-1}}(y)=\int\frac{1}{f'(x)}\,\cdot\,{dx} + c[itex]
This is only useful if the integral exists. In particular we need [itex] f'(x) [itex] to be non-zero across the range of integration.
It follows that functions with continuous derivative have inverses in a neighbourhood of every point where the derivative is non-zero. This need not be true if the derivative is not continuous.

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