ONE MORE STEP TOWARD BETTER TOMORROW : Differentiating Inverse Functions

Tuesday, 26 March 2013

Differentiating Inverse Functions

Inverse functions are very important in Mathematics as well as in many applied areas of science. The most famous pair of functions inverse to each other are the logarithmic and the exponential functions. Other functions like the tangent and arctangent play also a major role.
In any case, let f(x) be an invertible function, with f^-1(x) as its inverse, that is

$\begin{displaymath}f\circ f^{-1}(x) = x \;\;\mbox{and}\;\; f^{-1} \circ f(x) = x\;.\end{displaymath}$

May be the easiest way to remember this is to write

$\begin{displaymath}f(x) = y \;\;\mbox{iff}\;\; x = f^{-1}(y)\;.\end{displaymath}$

If f(x) is differentiable on an interval I, one may wonder whether f^-1(x) is also differentiable? The answer to this question hinges on f'(x) being equal to 0 or not . Indeed, if $f'(x) \neq 0$ for any $x \in I$ , then f^-1(x) is also differentiable. Moreover we have

$\begin{displaymath}\Big(f^{-1}\Big)'(x) = \frac{1}{f'\Big(f^{-1}(x)\Big)}\;\cdot \end{displaymath}$

Using Leibniz's notation, the above formula becomes

$\begin{displaymath}\frac{dx}{dy} = \frac{1}{dy/dx}\end{displaymath}$

which is easy to remember.Example. We will see in the coming pages that the logarithmic function $\ln(x)$ is the antiderivative of $\displaystyle \frac{1}{x}$ for x > 0, with $\ln(1) =0$ . This function has an inverse on $(0,\infty)$ , known to us as the exponential function e^x. So this function is differentiable and

$\begin{displaymath}\Big(e^x\Big)' = \frac{1}{\ln'(e^x)}\;\cdot\end{displaymath}$

But we have $\ln'(x) = \displaystyle \frac{1}{x}$ , hence

$\begin{displaymath}\Big(e^x\Big)' = \frac{1}{\displaystyle \frac{1}{e^x}} = e^x\;\cdot\end{displaymath}$

This is truly an amazing result: the derivative of e^x is the function itself. This property of the exponential function has many interesting applications.Example. Rational Powers. For x > 0, and any natural numbers n and m, we have

$\begin{displaymath}x^{n/m} = \Big(x^n\Big)^{1/m}.\end{displaymath}$

Let us first take care of the derivative of the function n^throot f(x)=x^1/n, which is just the inverse function of xⁿ. Clearly for x > 0, the derivative of xⁿ is not 0, so f(x)=x^1/n is differentiable and

$\begin{displaymath}\Big(x^{1/n}\Big)' = \frac{1}{n\Big(x^{1/n}\Big)^{n-1}}\cdot\end{displaymath}$

Easy algebraic calculations give

$\begin{displaymath}\Big(x^{1/n}\Big)' = \frac{1}{n} \frac{1}{x^{1-1/n}} = \frac{1}{n} x^{1/n -1}.\end{displaymath}$

In other words, the formula $\Big(x^r\Big)' = r x^{r-1}$ is also valid for r=1/n, for $n=1,2,\ldots$ Back to our formula, to differentiate the function x^n/m we will use the above results combined with the chain rule. So we have

$\begin{displaymath}\Big(x^{n/m}\Big)' = \Big(x^n\Big)' \frac{1}{m} \Big(x^n\Big)^{1/m -1} = \frac{n}{m} x^{n/m -1}\;.\end{displaymath}$

Again this means that the formula $\Big(x^r\Big)' = r x^{r-1}$ is valid even when r is a rational number. In particular combined with the chain rule, we get

$\begin{displaymath}\frac{d}{dx} \Big(u^{n/m}\Big) = \frac{n}{m} u^{n/m-1} \frac{du}{dx}\end{displaymath}$

for any differentiable function u(x).Example. We have

$\begin{displaymath}\frac{d}{dx} \Big((1+2x^3)^{1/5}\Big) = \frac{1}{5} (1+2x^3)^{-4/5}\; 6x^2 = \frac{6}{5} \frac{x^2}{(1+2x^3)^{4/5}}\end{displaymath}$

The following example discusses the above ideas for trigonometric functions.
Example. Consider the tangent function $\tan(x)$ on the interval $\left(\displaystyle -\frac{\pi}{2},\frac{\pi}{2}\right)$ . Then $\tan(x)$ has an inverse function $\tan^{-1}(x)$ known as the arctangent function, which we prefer to denote by $\arctan(x)$ . Since

$\begin{displaymath}\tan'(x) = \sec^2(x)= 1+ \tan^2(x) \neq 0\end{displaymath}$

for any $x \in \left(\displaystyle -\frac{\pi}{2},\frac{\pi}{2}\right)$ , then $\arctan(x)$ is differentiable. Moreover we have

$\begin{displaymath}\arctan'(x) = \frac{1}{\tan'\Big(\arctan(x)\Big)} = \frac{1}{1 + \tan^2\Big(\arctan(x)\Big)}\end{displaymath}$

$\begin{displaymath}\arctan'(x) = \frac{1}{1 + x^2}\cdot\end{displaymath}$

It might be surprising that the transcendental function $\arctan(x)$ has as its derivative a rational function. This will be very useful once we try to integrate rational functions.

ONE MORE STEP TOWARD BETTER TOMORROW

Tuesday, 26 March 2013

Differentiating Inverse Functions

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