Question

Differentiate $ta{n}^{-1}\left(\frac{1+2x}{1-2x}\right)$ with respect to $\sqrt{1+4x^2}$.

Answer 1

$\frac{d}{dx}{\tan }^{-1}\left(\frac{1+2x}{1-2x}\right)\frac{1}{1+{\left(\frac{1+2x}{1-2x}\right)}^2}\left[\frac{(1-2x)\left(2\right)-(1+2x)(-2)}{{(1-2x)}^2}\right]\frac{{(1-2x)}^2}{{(1+2x)}^2+{(1-2x)}^2}\left[\frac{2-4x+2+4x}{{(1-2x)}^2}\right]\frac{4}{1+4x^2+4x+1+{4x}^2-4x}=\frac{4}{2(1+4x^2)}=\frac{2}{1+4x^2}\frac{d}{dx}\left(\sqrt{1+4x^2}\right)=\frac{1}{2\sqrt{1+4x^2}}\left(8x\right)=\frac{4x}{\sqrt{1+4x^2}}\frac{d\left({\tan }^{-1}\left(\frac{1+2x}{1-2x}\right)\right)}{d\left(\sqrt{1+4x^2}\right)}=\frac{\frac{2}{1+4x^2}}{\frac{4x}{\sqrt{1+{4x}^2}}}=\frac{2}{1+{4x}^2}\times \frac{\sqrt{1+4x^2}}{4x}=\frac{1}{2x\sqrt{1+{4x}^2}}$