inverse functions and differentiation

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

:\left ^\right(y)=\frac.

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

:\mathcal\left ^\right\frac,

where $\mathcal$ 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,

:$\frac\,\cdot\, \frac = 1.$

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

:$\frac\,\cdot\, \frac = \frac$

considering that the derivative of with respect to ' is 1.

Derivation

Let $f$ be an invertible (bijective) function, let $x$ be in the domain of $f$, and let $y=f(x).$ Let $g=f^.$ So, $f(g(y))=y.$ Derivating this equation with respect to , and using the chain rule, one gets
:$f'(g(y))\cdot g'(y)=1.$
That is,
:$g'(y)=\frac 1$
or
:$(f^)^(y) = \frac.$

Examples

* $y = x^2$ (for positive ) 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.

* $y = e^x$ (for real ) has inverse $x = \ln$ (for positive $y$)

:$\frac = e^x \mbox\mbox\mbox\mbox; \mbox\mbox\mbox\mbox \frac = \frac = e^$

:$\frac\,\cdot\,\frac = e^x \cdot e^ = 1.$

Additional properties

* Integrating this relationship gives

::$(x)=\int\frac\, + C.$

: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:

::$\int f^(x)\, = x f^(x) - F(f^(x)) + C$

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

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

:$f'(x) = \frac = \frac \frac = \frac f''(x) \Rightarrow \frac = \frac$Therefore:

:$\frac = \frac = \frac\frac = \frac\frac = \frac$

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

:$\frac = \frac$

Higher derivatives

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

:$\frac\,\cdot\,\frac + \frac \left(\frac\right)\,\cdot\,\left(\frac\right) = 0,$

that is simplified further by the chain rule as

:$\frac\,\cdot\,\frac + \frac\,\cdot\,\left(\frac\right)^2 = 0.$

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

:$\frac = - \frac\,\cdot\,\left(\frac\right)^3.$

Similarly for the third derivative:

:$\frac = - \frac\,\cdot\,\left(\frac\right)^4 - 3 \frac\,\cdot\,\frac\,\cdot\,\left(\frac\right)^2$

or using the formula for the second derivative,

:$\frac = - \frac\,\cdot\,\left(\frac\right)^4 + 3 \left(\frac\right)^2\,\cdot\,\left(\frac\right)^5$

These formulas are generalized by the Faà di Bruno's formula.

These formulas can also be written using Lagrange's notation. If ' and ' are inverses, then

:$g''(x) = \frac$

Example

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

:$\frac = \frac = e^x = y \mbox\mbox\mbox\mbox; \mbox\mbox\mbox\mbox \left(\frac\right)^3 = y^3;$

so that

:$\frac\,\cdot\,y^3 + y = 0 \mbox\mbox\mbox\mbox; \mbox\mbox\mbox\mbox \frac = -\frac$,

which agrees with the direct calculation.

See also

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References

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{{Calculus topics

Articles containing proofs
Differentiation rules
Inverse functions
Theorems in mathematical analysis
Theorems in calculus