Inverse function rule

[[File:Umkehrregel 2.png|thumb|right|250px|The thick blue curve and the thick red curves are inverse to each other. A thin curve is the derivative of the same colored thick curve.

Inverse function 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 calculus, the inverse function rule is a formula that expresses the derivative of the inverse of a bijective and differentiable function $f$ in terms of the derivative of $f$. More precisely, if the inverse of $$f$$ is denoted as $$f^{-1}$$, where $$f^{-1}(y) = x$$ if and only if $$f(x) = y$$, then the inverse function rule is, in Lagrange's notation,


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

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$$f'(f^{-1}(a)) \ne 0$$. The same formula is also equivalent to the expression


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

where $$\mathcal{D}$$ 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.

The inverse function rule may also be expressed in Leibniz's notation. As that notation suggests,


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

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.

Derivation
Let $$f$$ be an invertible (bijective) function, let $$x$$ be in the domain of $$f$$, and let $$y$$ be in the codomain of $$f$$. Since f is a bijective function, $$y$$ is in the range of $$f$$. This also means that $$y$$ is in the domain of $$f^{-1}$$, and that $$x$$ is in the codomain of $$f^{-1}$$. Since $$f$$ is an invertible function, we know that $$f(f^{-1}(y)) = y$$. The inverse function rule can be obtained by taking the derivative of this equation.



\dfrac{\mathrm{d}}{\mathrm{d}y} f(f^{-1}(y)) = \dfrac{\mathrm{d}}{\mathrm{d}y} y $$

The right side is equal to 1 and the chain rule can be applied to the left side:



\begin{align} \dfrac{\mathrm{d}\left( f(f^{-1}(y)) \right)}{\mathrm{d}\left( f^{-1}(y) \right)} \dfrac{\mathrm{d}\left(f^{-1}(y)\right)}{\mathrm{d}y} &= 1 \\ \dfrac{\mathrm{d}f(f^{-1}(y))}{\mathrm{d}f^{-1}(y)} \dfrac{\mathrm{d}f^{-1}(y)}{\mathrm{d}y} &= 1 \\ f^{\prime}(f^{-1}(y)) (f^{-1})^{\prime}(y) &= 1 \end{align} $$

Rearranging then gives



(f^{-1})^{\prime}(y) = \frac{1}{f^{\prime}(f^{-1}(y))} $$

Rather than using $$y$$ as the variable, we can rewrite this equation using $$a$$ as the input for $$f^{-1}$$, and we get the following:



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

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} = e^{-x} $$


 * $$ \frac{dy}{dx}\,\cdot\,\frac{dx}{dy} = e^x \cdot e^{-x} = 1. $$

Additional properties

 * 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.


 * 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 $$ f $$.

Let $$ z = f'(x)$$  then we have, assuming $$ f(x) \neq 0$$:$$ \frac{d(f')^{-1}(z)}{dz} = \frac{1}{f(x)}$$This can be shown using the previous notation $$ y = f(x)$$. Then we have:
 * The inverse of the derivative of f(x) is also of interest, as it is used in showing the convexity of the Legendre transform.


 * $$ f'(x) = \frac{dy}{dx} = \frac{dy}{dz} \frac{dz}{dx} = \frac{dy}{dz} f(x) \Rightarrow \frac{dy}{dz} = \frac{f'(x) }{f(x)}$$Therefore:


 * $$ \frac{d(f')^{-1}(z)}{dz} = \frac{dx}{dz} = \frac{dy}{dz}\frac{dx}{dy} = \frac{f'(x)}{f(x)}\frac{1}{f'(x)} = \frac{1}{f(x)}$$

By induction, we can generalize this result for any integer $$ n \ge 1$$, with $$ z = f^{(n)}(x)$$, the nth derivative of f(x), and  $$ y = f^{(n-1)}(x)$$, assuming $$ f^{(i)}(x) \neq 0 \text{ for } 0 < i \le n+1 $$:


 * $$ \frac{d(f^{(n)})^{-1}(z)}{dz} = \frac{1}{f^{(n+1)}(x)}$$

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


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

that is simplified further by the chain rule as


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

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


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

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.