Question

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

• A. $\frac{1}{4t\sqrt{t}}$
• B. $\frac{\log_e 3}{4t\sqrt{t}}$
• C. ${2t^2}$
• D. $\frac{2t^2}{e^{\frac{1}{2}\log_e t}}$

Answer 1

Answer: (B)

$\frac{\log_e 3}{4t\sqrt{t}}$

Answer 2

Step 1: Simplify $y$.
The given equation is:
$y = e^{\frac{1}{2} \log_e t}$.
Using logarithmic properties:
$y = t^{\frac{1}{2}} = \sqrt{t}$.
Step 2: Simplify $x$.
The given equation is:
$x = \log_3 (e^t)$.
Using $\log_a (b^c) = c \cdot \log_a b$:
$x = t \cdot \log_3 e$.
Step 3: Differentiate $y$ with respect to $t$.
$\frac{dy}{dt} = \frac{d}{dt} (t^{\frac{1}{2}}) = \frac{1}{2} t^{-\frac{1}{2}} = \frac{1}{2\sqrt{t}}$.
Step 4: Differentiate $x$ with respect to $t$.
$\frac{dx}{dt} = \frac{d}{dt} (t \cdot \log_3 e) = \log_3 e$.
Step 5: Compute $\frac{dy}{dx}$.
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{2\sqrt{t}}}{\log_3 e}$.
Simplify:
$\frac{dy}{dx} = \frac{1}{2\sqrt{t} \cdot \log_3 e}$.
Using the property $\log_3 e = \frac{1}{\log_e 3}$, rewrite:
$\frac{dy}{dx} = \frac{\log_e 3}{2\sqrt{t}}$.
Final Answer:
$\frac{\log_e 3}{2\sqrt{t}}$