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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


(f^{-1})'(x)=\frac{1}{1+x^2}

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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