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. \( \frac{2t}{e^{2t}(4t - 1)} \)

Hint:

When simplifying logarithmic expressions, always use properties such as \( \log_e(a^b) = b \log_e(a) \) to simplify the equation. Also, remember that \( \log_e(e^z) = z \), which simplifies logarithmic expressions with exponents. In cases where you have an exponential function like \( t = e^{2x} \), this property makes differentiation much easier. Finally, remember that the derivative of a linear function like \( 4x \) is a constant, and its second derivative will be zero.

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.

Answer 2

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

Using the logarithmic property \( \log_e(a^b) = b \log_e(a) \), we can simplify \( \log_e(t^2) \) as:

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

Step 1: Substitute the value of \( t \):

We are given that \( t = e^{2x} \), so we substitute \( t \) into the expression:

\[ y = 2 \log_e(e^{2x}) \]

Step 2: Apply the property of logarithms:

We know that \( \log_e(e^z) = z \), so applying this property gives:

\[ y = 2 \cdot 2x = 4x \]

Step 3: Find the first derivative of \( y \) with respect to \( x \):

Now that we have \( y = 4x \), we can differentiate it with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}(4x) = 4 \]

Step 4: Find the second derivative of \( y \) with respect to \( x \):

The first derivative is a constant, so the second derivative is:

\[ \frac{d^2y}{dx^2} = 0 \]

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