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.

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.

Writing explicitly the dependence of y on x and the point at which the differentiation takes place and using Lagrange's notation, the formula for the derivative of the inverse becomes

<math>\left[f^{-1}\right]'(a)=\frac{1}{f'\left[f^{-1}(a)\right]}.</math>

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 \cdot\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}\,\cdot\,\frac{dx}{dy} = e^x \cdot \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.

See also

calculus, inverse functions, chain rule, inverse function theorem, implicit function theorem.