Inverse functions and differentiation

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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.

Their two derivatives, assuming they exist, are reciprocal, as the Leibniz notation suggests; 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.

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

$\left[f^{-1}\right]'(a)=\frac{1}{f'\left[ f^{-1}(a) \right]}.$

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.

Assuming that f has an inverse in a neighbourhood of x and that its derivative at that point is non-zero, its inverse is guaranteed to be differentiable at x and have a derivative given by the above formula.

1 Examples
2 Additional properties
3 Higher derivatives
4 Example
5 See also

[edit] Examples

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

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

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

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 $y$ )

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

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

[edit] 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.

[edit] Higher derivatives

The chain rule given above is obtained by differentiating the identity x=f ^-1(f(x)) with respect to x. One can continue the same process for higher derivatives. Differentiating the identity with respect to x two times, one obtains

$\frac{d^2y}{dx^2}\,\cdot\,\frac{dx}{dy} + \frac{d^2x}{dy^2}\,\cdot\,\left(\frac{dy}{dx}\right)^2 = 0$

or replacing the first derivative using the formula above,

$\frac{d^2y}{dx^2} = - \frac{d^2x}{dy^2}\,\cdot\,\left(\frac{dy}{dx}\right)^3$ .

Similarly for the third derivative:

$\frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\cdot\,\left(\frac{dy}{dx}\right)^4 - 3 \frac{d^2x}{dy^2}\,\cdot\,\frac{d^2y}{dx^2}\,\cdot\,\left(\frac{dy}{dx}\right)^2$

or using the formula for the second derivative,

$\frac{d^3y}{dx^3} = - \frac{d^3x}{dy^3}\,\cdot\,\left(\frac{dy}{dx}\right)^4 + 3 \left(\frac{d^2x}{dy^2}\right)^2\,\cdot\,\left(\frac{dy}{dx}\right)^5$