# Inverse functions and differentiation

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

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

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$f'(f^{-1}(a))\neq 0$ . 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.

## 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 $y$ )
${\frac {dy}{dx}}=e^{x}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac {dx}{dy}}={\frac {1}{y}}$ ${\frac {dy}{dx}}\,\cdot \,{\frac {dx}{dy}}=e^{x}\cdot {\frac {1}{y}}={\frac {e^{x}}{e^{x}}}=1$ ## Additional properties

${f^{-1}}(x)=\int {\frac {1}{f'({f^{-1}}(x))}}\,{dx}+C.$ 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:
$\int f^{-1}(x)\,{dx}=xf^{-1}(x)-F(f^{-1}(x))+C$ 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

${\frac {d^{2}y}{dx^{2}}}\,\cdot \,{\frac {dx}{dy}}+{\frac {d}{dx}}\left({\frac {dx}{dy}}\right)\,\cdot \,\left({\frac {dy}{dx}}\right)=0,$ that is simplified further by the chain rule as

${\frac {d^{2}y}{dx^{2}}}\,\cdot \,{\frac {dx}{dy}}+{\frac {d^{2}x}{dy^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{2}=0.$ Replacing the first derivative, using the identity obtained earlier, we get

${\frac {d^{2}y}{dx^{2}}}=-{\frac {d^{2}x}{dy^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{3}.$ Similarly for the third derivative:

${\frac {d^{3}y}{dx^{3}}}=-{\frac {d^{3}x}{dy^{3}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{4}-3{\frac {d^{2}x}{dy^{2}}}\,\cdot \,{\frac {d^{2}y}{dx^{2}}}\,\cdot \,\left({\frac {dy}{dx}}\right)^{2}$ or using the formula for the second derivative,

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

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

which agrees with the direct calculation.