Question

Find $\frac{dy}{dx}$ if $y={\tan }^{-1}\left(\frac{4x}{1+5x^2}\right)$.

Answer 1

We have,

$y={\tan }^{-1}\frac{4x}{1+5x^2}$

Differentiating w.r.t. $x$, we have

$\frac{dy}{dx}=\frac{1}{1+{\left(\frac{4x}{1+5x^2}\right)}^2}\left(\frac{4(1+5x^2)-10x\cdot 4x}{{(1+5x^2)}^2}\right)$

$\Rightarrow \frac{dy}{dx}=\left(\frac{{(1+5x^2)}^2}{(1+25x^4+10x^2)+16x^2}\right)\left(\frac{4+20x^2-40x^2}{{(1+5x^2)}^2}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{4-20x^2}{25x^4+26x^2+1}$

$\Rightarrow \frac{dy}{dx}=\frac{4(1-5x^2)}{(x^2+1)(25x^2+1)}$

Hence, this is the answer.