Question

Let $y = t^{10} + 1$ and $x = t^8 + 1$, then $\frac{d^{2}y}{dx^{2}}$ is equal to

• A. $\frac{5}{2}t$
• B. $20t^8$
• C. $\frac{5}{16t^6}$
• D. None of these

Answer 1

Answer: (C)

We have,
$y = t^{10} + 1$,
$x = t^8 + 1$
$\Rightarrow \frac{dy}{dt}=10t^{9}$,
$\frac{dx}{dt}=8t^{7}$
$\therefore \frac{dy}{dx}=\frac{dy / dt}{dx / dt}$
$=\frac{10t^{9}}{8t^{7}}$$=\frac{5}{4}t^{2}$
$\Rightarrow \frac{d^{2}\,y}{dx^{2}}=\frac{5}{4}\left(2t\right) \frac{dt}{dx}$
$=\frac{5}{4}\times2t \times\frac{1}{8t^{7}}$
$=\frac{5}{16t^{6}} $