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

Answer: (A)

$0$

Answer 2

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.