If \( y=e^{2\log_{e}t} \) and \( x=\log_{e}e^{t^{2}} \), then the value of \( \displaystyle\frac{d^{2}y}{dx^{2}} \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Since \( e^{\log_e u}=u \) and \( \log_e e^u=u \), the exponential and logarithm cancel in both expressions:
\( \displaystyle y=e^{2\log_e t}=e^{\log_e t^2}=t^2 \)
\( \displaystyle x=\log_e e^{t^2}=t^2 \)
So y = x, a straight line:
\( \displaystyle \frac{dy}{dx}=1 \)
\( \displaystyle \frac{d^2y}{dx^2}=0 \)