Question

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

Answer 1

$y=ta{n}^{-1}\left(\frac{1}{x^2+x+1}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{1}{1+(\frac{1}{x^2+x+1}{)}^2}\times \left[\frac{1\times (2x+1)}{(x^2+x+1{)}^2}\right]$

$\Rightarrow \frac{dy}{dx}=-\frac{(2x+1)}{1+(x^2+x+1{)}^2}$

$\frac{dy}{dx}=\frac{-(2x+1)}{x^4+x^2+1+2x^3+2x+2x^2+1}$

$\frac{dy}{dx}=\frac{-(2x+1)}{x^4+2x^3+3x^2+2x+2}$