Question

If $t = e^{2x}$ and $y = \ln(t^2)$, then $\frac{d^2 y}{dx^2}$ is:

• A. $0$
• B. $4t$
• C. \( \frac{2t}{4e^t} \)
• D. \( \frac{2t}{e^{2t}(4t - 1)} \)

Answer 1

The Correct Option is A

First, simplify \( y = \log_e(t^2) \) as follows:

\( y = 2 \log_e(t) \)

Since \( t = e^{2x} \), we have:

\( \log_e(t) = 2x \implies y = 2 \cdot 2x = 4x \)

Now, taking the first derivative with respect to \( x \):

\( \frac{dy}{dx} = 4 \)

Then, taking the second derivative with respect to \( x \):

\( \frac{d^2y}{dx^2} = 0 \)

Thus, the value of \( \frac{d^2y}{dx^2} \) is 0.