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

$x=log(1+t^2),y=t-ta{n}^{-1}t$

To show $\frac{dy}{dx}=\sqrt{\frac{e^{x-1}}{2}}$

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

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

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

y = t-tan t

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

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

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

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

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

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

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

$t^2=e^x-1$

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

Put in (1)

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

Hence proved