Inverse functions and differentiation

Short description: Calculus identity

Rule:
$f^{\prime }(x)=\frac{1}{(f^{-1}{)}^{\prime }(f(x))}$

Example for arbitrary $x_0\approx 5.8$:
$f^{\prime }(x_0)=\frac{1}{4}$
$(f^{-1}{)}^{\prime }(f(x_0))=4$

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 as $f^{-1}$, where $f^{-1}(y)=x$ if and only if $f(x)=y$.

Their two derivatives, assuming they exist, are reciprocal, as the Leibniz notation suggests; that is:

$\frac{dx}{dy}\cdot \frac{dy}{dx}=1.$

This relation is obtained by differentiating the equation $f^{-1}(y)=x$ in terms of x and applying the chain rule, yielding that:

$\frac{dx}{dy}\cdot \frac{dy}{dx}=\frac{dx}{dx}$

considering that 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, the formula for the derivative of the inverse becomes (in Lagrange's notation):

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

This formula holds in general whenever $f$ is continuous and injective on an interval I, with $f$ being differentiable at $f^{-1}(a)$($\in I$) and where$f^{\prime }(f^{-1}(a))\ne 0$. The same formula is also equivalent to the expression

${𝒟}\left[f^{-1}\right]=\frac{1}{({𝒟}f)\circ \left(f^{-1}\right)},$

where ${𝒟}$ denotes the unary derivative operator (on the space of functions) and $\circ$ denotes function composition.

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 $x=\sqrt{y}$.

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

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

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 $y$)

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

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

Additional properties

• Integrating this relationship gives

$f^{-1}(x)=\int \frac{1}{f^{\prime }(f^{-1}(x))}dx+C.$

This is only useful if the integral exists. In particular we need $f^{\prime }(x)$ 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.

• Another very interesting and useful property is the following:

$\int f^{-1}(x)dx=x{f}^{-1}(x)-F(f^{-1}(x))+C$

Where $F$ denotes the antiderivative of $f$.

Higher derivatives

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

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

that is simplified further by the chain rule as

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

Replacing the first derivative, using the identity obtained earlier, we get

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

Similarly for the third derivative:

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

or using the formula for the second derivative,

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

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

$g^{″}(x)=\frac{-f^{″}(g(x))}{[f^{\prime }(g(x)){]}^3}$

Example

• $y=e^x$ has the inverse $x=\ln y$. Using the formula for the second derivative of the inverse function,

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

so that

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

which agrees with the direct calculation.

See also

• Calculus
• Inverse functions
• Chain rule
• Inverse function theorem
• Implicit function theorem
• Integration of inverse functions