Question

If $x=log(1+t^2)$ and $y=t-ta{n}^{-1}t$. Then, $\frac{dy}{dx}$ is equal to

• A. $e^x-1$
• B. $t^2-1$
• C. $\frac{\sqrt{e^x-1}}{2}$
• D. $e^x-y$

Answer 1

The correct option is C $\frac{\sqrt{e^x-1}}{2}$

Differentiating wrt 't'

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

$\frac{dy}{dt}=\frac{d}{dt}(t-{\tan }^{-1}t)=1-\frac{1}{1+t^2}=\frac{t^2}{1+t^2}$

$\frac{dy}{dx}=\frac{t^2/(1+t^2)}{2t/(1+t^2)}=\frac{t}{2}$

$x=log(1+t^2)$

$e^x=1+t^2$

$t=\sqrt{e^x-1}$

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

$\therefore \frac{dy}{dx}=\frac{\sqrt{e^x-1}}{2}$.