If \(t=e^{2x}\) and \(y=\log_e t^2\), then \(\displaystyle\frac{d^2y}{dx^2}\) is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Since \(t = e^{2x} \gt 0\), the law of logarithms gives
\(y = \log_e t^2 = 2\log_e t\)
and \(\log_e t = \log_e e^{2x} = 2x\). So y is simply a linear function of x:
\(y = 2 \times 2x = 4x\)
\( \displaystyle \frac{dy}{dx} = 4 \)
\( \displaystyle \frac{d^2y}{dx^2} = 0 \)