Question

$\frac{d}{dx}{\tan }^{-1}\left[\frac{x^{1/3}+a^{1/3}}{1-x^{1/3}{a}^{1/3}}\right]$.

Answer 1

Let $y={\tan }^{-1}\left[\frac{x^{1/3}+a^{1/3}}{1-x^{1/3}{a}^{1/3}}\right]$

now, we know that

${\tan }^{-1}\left(\frac{a+b}{1-a.b}\right)={\tan }^{-1}a+{\tan }^{-1}b.$

So, $y={\tan }^{-1}\left(x^{1/3}\right)+{\tan }^{-1}\left(a^{1/3}\right)$

here a is constant i.e. ${\tan }^{-1}\left(a^{1/3}\right)$ is also constant.

$\frac{dy}{dx}=\frac{d}{dx}\left({\tan }^{-1}\left[\frac{x^{1/3}+a^{1/3}}{x-x^{1/3}{a}^{1/3}}\right]\right)$

$=\frac{d}{dx}\left({\tan }^{-1}\right(x^{1/3})-{\tan }^{-1}(a^{1/3}\left)\right)$

$=\frac{d}{dx}\left({\tan }^{-1}\right(x^{1/3}\left)\right)-0$.

now, $\frac{d}{dx}\left({\tan }^{-1}A\right)=\frac{1}{1+A^2}.\frac{dA}{dx}$

So, $\frac{dy}{dx}$$=\frac{1}{1+\left(x^{1/3}\right)}$$\times \left(\frac{1}{3}\right).$$(x{)}^{1/3-2}$

$=\frac{1}{3}\left(\frac{1}{1+x^{2/3}}\right)(x{)}^{-2/3}$

$\frac{d}{dx}{\tan }^{-1}\left[\frac{x^{1/3}+a^{1/3}}{1-x^{1/3}{a}^{1/2}}\right]=\frac{1}{3}\left(\frac{1}{x^{2/3}+x^{4/3}}\right)$