# Inverse functions and differentiation Rule:
${\color }(x)={\frac {{\color {(f^{-1})'}}({\color }(x))}}$ Example for arbitrary $x_\approx 5.8$ :
${\color }(x_)={\frac }$ ${\color {(f^{-1})'}}({\color }(x_))=4~$ 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:

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

${\frac }\,\cdot \,{\frac }={\frac }$ 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):

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

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

${\mathcal }\left[f^{-1}\right]={\frac {({\mathcal }f)\circ \left(f^{-1}\right)}},$ where ${\mathcal }\!$ denotes the unary derivative operator (on the space of functions) and $\circ \!$ denote the binary composition operator.

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

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

• $\,y=e^$ (for real x) has inverse $\,x=\ln$ (for positive $y$ )
${\frac }=e^{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac }={\frac }$ ${\frac }\,\cdot \,{\frac }=e^\cdot {\frac }={\frac }}}=1$ • Integrating this relationship gives
$}(x)=\int {\frac }(x))}}\,+C.$ 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 $f^{-1}(f(x))=x\!$ with respect to x. One can continue the same process for higher derivatives. Differentiating the identity twice with respect to x, one obtains

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

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

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

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

${\frac y}}}=-{\frac x}}}\,\cdot \,\left({\frac }\right)^+3\left({\frac x}}}\right)^\,\cdot \,\left({\frac }\right)^$ 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

$g''(x)={\frac {-f''(g(x))}{[f'(g(x))]^}}$ ## Example

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

${\frac x}}}\,\cdot \,y^+y=0{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }};{\mbox{ }}{\mbox{ }}{\mbox{ }}{\mbox{ }}{\frac x}}}=-{\frac }}$ ,

which agrees with the direct calculation.