If \(x = t^{2}\) and \(y = \log_e t\), where \(t \gt 0\), then \(\displaystyle\frac{d^{2}y}{dx^{2}}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(-\displaystyle\frac{1}{2t^{4}}\)
The parametric derivative is
\( \displaystyle \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1/t}{2t} = \frac{1}{2t^{2}} \)
For the second derivative, \(\frac{1}{2t^{2}}\) is differentiated with respect to x, again through t:
\( \displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d}{dt}\left(\frac{1}{2t^{2}}\right) \cdot \frac{dt}{dx} = -\frac{1}{t^{3}} \cdot \frac{1}{2t} = -\frac{1}{2t^{4}} \)
A quick check: \(t = \sqrt{x}\), so \(y = \frac{1}{2}\log_e x\) and \(y'' = -\frac{1}{2x^{2}} = -\frac{1}{2t^{4}}\).
Why the other options are incorrect
Option (1) \(-\frac{1}{t^{3}}\) — Incorrect. This is only \(\frac{d}{dt}\left(\frac{1}{2t^{2}}\right)\); it still has to be multiplied by \(\frac{dt}{dx} = \frac{1}{2t}\).
Option (2) \(\frac{1}{2t^{2}}\) — Incorrect. This is the first derivative dy/dx, not the second.
Option (3) \(-\frac{1}{2t^{2}}\) — Incorrect. This is \(\frac{d^{2}y/dt^{2}}{d^{2}x/dt^{2}} = \frac{-1/t^{2}}{2}\), which is not the second derivative.