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
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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