Find the $n^{th}$ derivative of $y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$
My Attempt:
$$y=\tan^{-1} \left(\dfrac {1+x}{1-x}\right)$$
Put $x=\cos (\theta)$
$$y=\tan^{-1} \left(\dfrac {1+\cos (\theta)}{1-\cos (\theta)}\right)$$
$$y=\tan^{-1} \left(\dfrac {2\cos^2 \left(\dfrac {\theta}{2}\right)}{2\sin^2\left(\dfrac {\theta}{2}\right)}\right)$$
$$y=\tan^{-1} \left(\cot^2 \left(\dfrac {\theta}{2}\right)\right)$$
Best Answer
Once you find the first derivative, which is $$ f'(x)=\frac{1}{x^2+1} $$ your question reduces to the following one:
What is the $n^\text{th}$ derivative of $f(x)=\frac{1}{1+x^2}$
or more generally, this one:
How find this $\left(\frac{1}{x^2+a^2}\right)^{(n)}$
To find the first derivative, you can simply apply the chain rule.