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.

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.

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Examples

\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.

\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

Additional properties

{f^{-1}}(x)=\int\frac{1}{f'({f^{-1}}(x))}\,{dx} %2B 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.

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} %2B \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 %2B 3 \left(\frac{d^2x}{dy^2}\right)^2\,\cdot\,\left(\frac{dy}{dx}\right)^5

These formulas are generalized by the Faà di Bruno's formula.

These formulas can also be written using Lagrange's notation. If f and g are inverses, then

g''(x) = \frac{-f''(g(x))}{[f'(g(x))]^3}

Example

\frac{dy}{dx} = \frac{d^2y}{dx^2} = e^x = y \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \left(\frac{dy}{dx}\right)^3 = y^3;

so that

\frac{d^2x}{dy^2}\,\cdot\,y^3 %2B y = 0 \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{d^2x}{dy^2} = -\frac{1}{y^2} ,

which agrees with the direct calculation.

See also