Question

If $x=\log (1+t^2),y=t-{\tan }^{-1}t$, show that $\frac{dy}{dx}=\frac{\sqrt{e^x-1}}{2}$

Answer 1

Given,

$y=t-{\tan }^{-1}t$

$\frac{dy}{dt}=\frac{d}{dt}(t-{\tan }^{-1}t)$

$\therefore \frac{dy}{dt}=1-\frac{1}{t^2+1}$

$x=\log (1+t^2)\Rightarrow e^x-1=t^2\Rightarrow t=\sqrt{e^x-1}$

$\frac{dx}{dt}=\frac{d}{dt}\left(\log \right(1+t^2\left)\right)$

$\therefore \frac{dx}{dt}=\frac{2t}{t^2+1}$

$\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$=\frac{1-\frac{1}{t^2+1}}{\frac{2t}{t^2+1}}$

$=\frac{t^2}{2t}$

$=\frac{t}{2}$

$=\frac{\sqrt{e^x-1}}{2}$