Excercise 4
The inverse function of the tangent is denoted by \(\arctan\) : \(\mathbb{R} \rightarrow (-\frac{\pi}{2},\frac{\pi}{2}).\)
\((a) \)Find the derivative of \(\arctan\) using the rules for differentiation of the inverse function.
\((b)\) The function \(\arctan\) has a power series expansion for all \(x\in \mathbb{R}\) with \(\mid x\mid < 1:\)
\(\arctan x=\sum_{n=0}^{+\infty}(-1)^{n}\frac{x^{2n+1}}{2n+1}.\)
Calculate \((\arctan x)' \)by differentiating this power series.
Solution
It applies
\((\arctan x)'=\left. \frac{1}{(\tan y)'}\right|_{y=\arctan x}.\)
It is \((\tan y)'= (\cos 2y)-1\). Using the identity undefined \(1= \cos^{2}y+\sin^{2}y\) we receive
\(\frac{1}{(\tan y)'} =\cos^{2}y=\frac{\cos^{2}y}{\sin^{2} y +\cos^{2} y} =\frac{1}{\tan^{2} y+1} .\)
Also
\(\left. (\arctan x)' =\frac{1}{\tan^{2} y+1}\right| _{y=\arctan x}=\dfrac{1}{1 + x^{2}} .\)