Question

Find $\frac{dy}{dx}$ if $y={\tan }^{-1}\left(\frac{5x+1}{3-x-6x^2}\right)$

Answer 1

According to the problem,

$\begin{array}{c}\frac{d}{dx}ta{n}^{-1}\left(\frac{5x+1}{3-x-x^2}\right)\\ \frac{d}{dx}\left({\tan }^{-1}\left(\frac{5x+1}{3-x-x^2}\right)\right)\end{array}$

Applying chain rule,

$\begin{array}{c}f={\tan }^{-1}\left(u\right),u=\frac{5x+1}{3-x-x^2}\\ =\frac{d}{du}\left({\tan }^{-1}\left(u\right)\right)\frac{d}{dx}\left(\frac{5x+1}{3-x-x^2}\right)\\ =\frac{1}{1+u^2}.\left(\frac{\left(5x+1\right)\left(-1-x\right)+\left(3-x-x^2\right)\left(5\right)}{{\left(3-x-x^2\right)}^2}\right)\\ =\frac{1}{{\left(\frac{5x+1}{3-x-x^2}\right)}^2+1}.\frac{5x^2+2x+16}{{\left(3-x-x^2\right)}^2}\\ =\frac{5x^2+2x+16}{x^4+2x^3+20x^2+4x+10}\end{array}$