Question:medium

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

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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.

Updated On: Mar 27, 2026
  • $0$
  • $4t$
  • \( \frac{2t}{4e^t} \)
  • \( \frac{2t}{e^{2t}(4t - 1)} \)
Show Solution

The Correct Option is A

Solution and Explanation

Simplify \( y = \log_e(t^2) \) to \( y = 2 \log_e(t) \).

Given \( t = e^{2x} \), it follows that \( \log_e(t) = 2x \). Substituting this into the expression for \( y \) yields \( y = 2 \cdot 2x = 4x \).

The first derivative with respect to \( x \) is \( \frac{dy}{dx} = 4 \).

The second derivative with respect to \( x \) is \( \frac{d^2y}{dx^2} = 0 \).

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

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