# Inverse functions and differentiation

Rule:
${\color{CornflowerBlue}{f'}}(x) = \frac{1}{{\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x))}$

Example for arbitrary $x_0 \approx 5.8$:
${\color{CornflowerBlue}{f'}}(x_0) = \frac{1}{4}$
${\color{Salmon}{(f^{-1})'}}({\color{Blue}{f}}(x_0)) = 4~$

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.

## Examples

• $\,y = x^2$ (for positive $x$) has inverse $x = \sqrt{y}$.
$\frac{dy}{dx} = 2x \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{dx}{dy} = \frac{1}{2\sqrt{y}}=\frac{1}{2x}$
$\frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = 2x \cdot\frac{1}{2x} = 1.$

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.

• $\,y = e^x$ (for real $x$) has inverse $\,x = \ln{y}$ (for positive $y$)
$\frac{dy}{dx} = e^x \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \frac{dx}{dy} = \frac{1}{y}$
$\frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = e^x \cdot \frac{1}{y} = \frac{e^x}{e^x} = 1$

• Integrating this relationship gives
${f^{-1}}(x)=\int\frac{1}{f'({f^{-1}}(x))}\,{dx} + c.$
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 a function that has a continuous derivative has an inverse 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. Twice differentiating the identity with respect to x 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,

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

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

• $\,y = e^x$ has the inverse $\,x = \ln y$. Using the formula for the second derivative of the inverse function,
$\frac{dy}{dx} = \frac{d^2y}{dx^2} = e^x = y \mbox{ }\mbox{ }\mbox{ }\mbox{ }; \mbox{ }\mbox{ }\mbox{ }\mbox{ } \left(\frac{dy}{dx}\right)^3 = y^3;$

so that

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

which agrees with the direct calculation.