inverse functions and differentiation

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.

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Inverse Functions And Differentiation

Examples

Additional properties

Related topics