Question

$y=co{t}^{-1}\left(\frac{1-3x^2}{3x-x^3}\right);\mid x\mid >\frac{1}{\sqrt{3}}$ ten evaluate $\frac{dy}{dx}$

Answer 1

We have,

$y={\cot }^{-1}\left(\frac{1-3x^2}{3x-x^3}\right);\left|x\right|>\frac{1}{\sqrt{3}}$

Put $x=\cot \theta \Rightarrow \theta ={\cot }^{-1}x$

Then,

Put and we get,

$y={\cot }^{-1}\left(\frac{1-3{\cot }^2\theta }{3\cot \theta -co{t}^3\theta }\right)$

$y={\tan }^{-1}\left(\frac{3\cot \theta -co{t}^3\theta }{1-3{\cot }^2\theta }\right)$

$={\tan }^{-1}\left(\frac{co{t}^3\theta -3\cot \theta }{3{\cot }^2\theta -1}\right)$

$={\tan }^{-1}\cot 3\theta$

$={\tan }^{-1}\tan (\frac{\pi }{2}-3\theta )$

$=\frac{\pi }{2}-3\theta$

$y=\frac{\pi }{2}-3{\cot }^{-1}x$

On differentiating and we get,

$\frac{dy}{dx}=\frac{d}{dx}(\frac{\pi }{2}-3{\cot }^{-1}x)$

$=(0-3(-\frac{1}{1+x^2}))$

$=\frac{3}{1+x^2}$

Hence, this is the answer.