Solving differential equations



		Site Map

Inverse functions

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.

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 Leibniz notation suggests, reciprocal, 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. 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

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

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

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

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.

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

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

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

Additional properties

Integrating this relationship gives

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

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.

Solving linear equations