The
inverse of a
function <math>y = f(x)</math> is a function that, in some fashion, "undoes" the effect of <math>f</math> (see
inverse function for a formal and detailed definition). The inverse of <math>f</math> is denoted <math>f^{-1}</math>. 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.
<math>\frac{dy}{dx} </math> denotes the derivative of the function <math>y=f(x)</math> with respect to <math>x</math>.
<math>\frac{dx}{dy} </math> denotes the derivative of the function <math>x=f(y)</math> with respect to <math>y</math>.
The two derivatives are, as the Leibnitz notation[?] suggests, reciprocal, that is
- <math>\frac{dx}{dy}\,.\, \frac{dy}{dx} = 1 </math>
This is a direct consequence of the chain rule, since
- <math> \frac{dx}{dy}\,.\, \frac{dy}{dx} = \frac{dx}{dx} </math>
and the derivative of <math> x </math> with respect to <math> x </math> is 1.
- <math>y = x^2</math> (for positive <math>x</math>) has inverse <math>x = \sqrt{y}</math>.
- <math> \frac{dy}{dx} = 2x
\mbox{ }\mbox{ }\mbox{ }\mbox{ };
\mbox{ }\mbox{ }\mbox{ }\mbox{ }
\frac{dx}{dy} = \frac{1}{2\sqrt{y}} </math>
- <math> \frac{dy}{dx}\,.\,\frac{dx}{dy} = 2x . \frac{1}{2\sqrt{y}} = \frac{2x}{2x} = 1 </math>
- <math>y = e^x</math> has inverse <math> x = \ln (y)</math> (for positive <math>y</math>).
- <math> \frac{dy}{dx} = e^x
\mbox{ }\mbox{ }\mbox{ }\mbox{ };
\mbox{ }\mbox{ }\mbox{ }\mbox{ }
\frac{dx}{dy} = \frac{1}{y} </math>
- <math> \frac{dy}{dx}\,.\,\frac{dx}{dy} = e^x . \frac{1}{y} = \frac{e^x}{e^x} = 1 </math>
- Integrating this relationship gives
- <math>{f^{-1}}(y)=\int\frac{1}{f'(x)}\,.\,{dx} + c</math>
- This is only useful if the integral exists. In particular we need <math> f'(x) </math> 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.
calculus, inverse functions, chain rule