Question

If $f\left(x\right)={\tan }^{-1}\left(\frac{3-x^3}{x+3x^2}\right),$ then value of $f^{\prime }(-1)$ is

Answer 1

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

$f^{\prime }\left(x\right)=\frac{1}{1+\frac{9}{x^2}}\times (-\frac{3}{x^2})-\frac{2x}{1+x^4}⟹f^{\prime }(-1)=-\frac{3}{10}+2=\frac{17}{10}$