Let y = tan-1 \(\frac{4\sqrt{x}}{1-4x}\)
Put 2√x = tanθ
differentiating w.r.t x
\(2.\frac{1}{2\sqrt{x}}=sec^2\theta\,\frac{d\theta}{dx}\)
⇒ \(\frac{d\theta}{dx} =\frac{cos^2 \theta}{\sqrt{x}}=\frac{1}{\sqrt{x}(1+4x)}\)
Also,
\(y=tan^{-1} \left(\frac{2.2\sqrt{x}}{1-(2\sqrt{x})^2}\right)\)
\(=tan^{-1}\left(\frac{2\,tan\,\theta}{1-tan^2\,\theta}\right)\)
\(=tan^ {-1}(tan\,2\theta)=2\theta\)
\(\therefore \frac{dy}{d\theta}=2\,and\,\frac{d\theta}{dx}=\frac{1}{\sqrt{x}(1+4x)}\)
\(\frac{dy}{dx}=\frac{dy}{d\theta}.\frac{d\theta}{dx}=2\times \frac{1}{\sqrt{x}(1+4x)}=\frac{2}{\sqrt{x}(1+4x)}\)
