If \( x=t^2 \) and \( y=t\sqrt{t} \), then \( \displaystyle\frac{d^2y}{dx^2} \) at \( t=\displaystyle\frac{1}{4} \) is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → −6
Here \( y=t^{3/2} \). Differentiating with respect to t:
\( \displaystyle \frac{dx}{dt}=2t \)
\( \displaystyle \frac{dy}{dt}=\frac{3}{2}t^{1/2} \)
\( \displaystyle \frac{dy}{dx}=\frac{\frac{3}{2}t^{1/2}}{2t}=\frac{3}{4}t^{-1/2} \)
For the second derivative, differentiate again with respect to t and divide by \( \frac{dx}{dt} \):
\( \displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{3}{4}t^{-1/2}\right)}{dx/dt} \)
\( \displaystyle =\frac{-\frac{3}{8}t^{-3/2}}{2t} \)
\( \displaystyle =-\frac{3}{16}t^{-5/2} \)
At \( t=\frac{1}{4} \), \( t^{-5/2}=4^{5/2}=32 \):
\( \displaystyle \frac{d^2y}{dx^2}=-\frac{3}{16}\times 32=-6 \)
Shortcut: Or eliminate t first: \( t=x^{1/2} \), so \( y=t^{3/2}=x^{3/4} \).
\( \displaystyle y''=\frac{3}{4}\cdot\left(-\frac{1}{4}\right)x^{-5/4}=-\frac{3}{16}x^{-5/4} \)
At \( t=\frac{1}{4} \), \( x=\frac{1}{16} \) and \( x^{-5/4}=16^{5/4}=32 \), so \( y''=-6 \).