Inverse functions and differentiation
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In mathematics, 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 Leibniz notation suggests, reciprocal, that is
- <math>\frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = 1 <math>
This is a direct consequence of the chain rule, since
- <math> \frac{dx}{dy}\,\cdot\, \frac{dy}{dx} = \frac{dx}{dx} <math>
and the derivative of <math> x <math> with respect to <math> x <math> 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
- <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}\,\cdot\,\frac{dx}{dy} = 2x . \frac{1}{2\sqrt{y}} = \frac{2x}{2x} = 1 <math>
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.
- <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>
Additional properties
- Integrating this relationship gives
- <math>{f^{-1}}(y)=\int\frac{1}{f'(x)}\,\cdot\,{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.
Related topics
calculus, inverse functions, chain rule, inverse function theorem, implicit function theorem.