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# Inverse functions and differentiation

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.

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

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

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

The two derivatives are, as the Leibnitz notation[?] suggests, reciprocal, that is

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

This is a direct consequence of the chain rule, since

$\frac{dx}{dy}\,.\, \frac{dy}{dx} = \frac{dx}{dx}$

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

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

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

$\frac{dy}{dx}\,.\,\frac{dx}{dy} = 2x . \frac{1}{2\sqrt{y}} = \frac{2x}{2x} = 1$

• $y = e^x$ has inverse $x = \ln (y)$ (for positive $y$).

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

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

• Integrating this relationship gives

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

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 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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