Calculating Derivative – Computing a derivative of an inverse function of tan – Exercise 2088

Exercise

Find the derivative of the inverse of the following function:

$f(x)=\tan x$

in the interval:

$x\in (-\frac{\pi}{2}, \frac{\pi}{2})$

Final Answer

Show final answer

Solution

Given the function:

$f(x)=\tan x$

Its inverse function is

$f^{-1}(x)=\arctan x$

We use the formula to find the derivative of the inverse function and get:

$(f^{-1})'(x)=(\arctan x)'=$

$=\frac{1}{(\tan (\arctan x))'}=$

$=\frac{1}{\frac{1}{cos^2 (\arctan x)}}=$

$=cos^2 (\arctan x)=$

Using the following trigonometric identity:

$\cos x = \frac{1}{\sqrt{1+\tan^2 x}}$

we get:

$= \frac{1}{1+\tan^2 (\arctan x)}=$

$= \frac{1}{1+x^2}$

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