If $y=e^{\frac{1}{2}\log_et}$ and $x=\log_3(e^{t^2})$, then $\frac{dy}{dx}$ is equal to:

$\frac{\log_e3}{4t\sqrt{t}}$

The correct answer is Option (3) → $\frac{\log_e3}{4t\sqrt{t}}$

$y=e^{\frac{1}{2}\log_et}$ 

$⇒y=t^{\frac{1}{2}}=\sqrt{t}$  $(∵e^{\log_ea})$

$⇒\frac{dy}{dt}=\frac{1}{2}t^{-\frac{1}{2}}=\frac{1}{2\sqrt{t}}$   ...(1)

Now,

$x=\log_3(e^{t^2})$

$x=t^2\log_3e$   $(∵\log_ab^c=c\log_ab)$

$x=\frac{t^2}{\log_e3}$   $\left(∵\log_e3=\frac{1}{\log_e3}\right)$

$⇒\frac{dx}{dt}=\frac{2t}{\log_e3}$   ...(2)

From (1) & (2)

$\frac{dy}{dx}=\frac{\frac{1}{2\sqrt{t}}}{\frac{2t}{\log_e3}}=\frac{1}{2\sqrt{t}}×\frac{\log_e3}{2t}=\frac{\log_e3}{4t\sqrt{t}}$