If \( x=\log_{e}t \) and \( y=\displaystyle\frac{1}{t} \), then \( \displaystyle\frac{d^{2}y}{dx^{2}} \) at \( t=0.1 \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 10
From \( x=\log_{e}t \), \( t=e^{x} \), so y can be written in terms of x.
\( \displaystyle y=\frac{1}{t} \)
\( \displaystyle =e^{-x} \)
Differentiating twice with respect to x,
\( \displaystyle \frac{dy}{dx}=-e^{-x} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}}=e^{-x} \)
\( \displaystyle =\frac{1}{t} \)
At t = 0.1,
\( \displaystyle \frac{d^{2}y}{dx^{2}}=\frac{1}{0.1} \)
\( \displaystyle =10 \)