Inverse functions and differentiation

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 $f^\{-1\}$. The statements "y=f(x)" and "x=f ^-1(y)" are equivalent.

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

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

This is a direct consequence of the chain rule, since

:$frac\{dx\}\{dy\},cdot,\; frac\{dy\}\{dx\}\; =\; frac\{dx\}\{dx\}$

and 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 and using Lagrange's notation, the formula for the derivative of the inverse becomes

:$left\; [f^\{-1\}\; ight]\; \text{'}(a)=frac\{1\}\{f\text{'}left\; [\; f^\{-1\}(a)\; ight]\; \}.$

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 $x\; =\; sqrt\{y\}$.

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

:$frac\{dy\}\{dx\},cdot,frac\{dx\}\{dy\}\; =\; 2x\; cdotfrac\{1\}\{2sqrt\{y\; =\; frac\{2x\}\{2x\}\; =\; 1.$

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.

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

:$frac\{dy\}\{dx\}\; =\; e^xmbox\{\; \}mbox\{\; \}mbox\{\; \}mbox\{\; \};mbox\{\; \}mbox\{\; \}mbox\{\; \}mbox\{\; \}frac\{dx\}\{dy\}\; =\; frac\{1\}\{y\}$

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

Additional properties

* Integrating this relationship gives

::$\{f^\{-1(y)=intfrac\{1\}\{f\text{'}(x)\},cdot,\{dy\}\; +\; c.$

:This is only useful if the integral exists. In particular we need $f\text{'}(x)$ to be non-zero across the range of integration.

:It follows that functions with continuous derivative have inverses 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 "x=f ^-1(f(x))" with respect to x. One can continue the same process for higher derivatives. Differentiating the identity with respect to x two times, one obtains

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

or replacing the first derivative using the formula above,

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

Similarly for the third derivative:

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

or using the formula for the second derivative,

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

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

Example

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

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

so that

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

which agrees with the direct calculation.

ee also

*calculus
*inverse functions
*chain rule
*inverse function theorem
*implicit function theorem