If \(y=e^{\frac{1}{2}\log_e t}\) and \(x=\log_3\left(e^{t^2}\right)\), then \(\displaystyle\frac{dy}{dx}\) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{\log_e 3}{4t\sqrt{t}}\)
Both functions simplify first. Since \(e^{\log_e a} = a\):
\(y = e^{\log_e t^{1/2}} = \sqrt{t}\)
Using \(\log_a b^c = c\log_a b\) and the change of base \(\log_3 e = \frac{1}{\log_e 3}\):
\( \displaystyle x = t^2\log_3 e = \frac{t^2}{\log_e 3} \)
Differentiating with respect to t:
\( \displaystyle \frac{dy}{dt} = \frac{1}{2\sqrt{t}},\quad \frac{dx}{dt} = \frac{2t}{\log_e 3} \)
For a parametric form:
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1}{2\sqrt{t}}\times\frac{\log_e 3}{2t} \)
\( \displaystyle = \frac{\log_e 3}{4t\sqrt{t}} \)