Question

Let $y=t^{10}+1$ and $x=t^8+1$, then $\frac{d^{2}y}{d{x}^2}$ 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$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{5}{4}\left(2t\right)\frac{dt}{dx}$
$=\frac{5}{4}\times 2t\times \frac{1}{8t^7}$
$=\frac{5}{16t^6}$