Question

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

• A. $\frac{5}{2}t$
• B. $20t^8$
• C. $\frac{5}{16t^6}$
• D. $\frac{15}{16t^6}$

Answer 1

The correct option is C $\frac{5}{16t^6}$

$y=t^{10}+1$

$\frac{dy}{dt}=10t^9$

$x=t^8+1$

$\frac{dx}{dt}=8t^7$

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{10t^9}{8t^7}=\frac{10}{8}{t}^2$

$\therefore \frac{d^{2}y}{d{x}^2}=\frac{10}{8}\frac{d}{dx}\left(t^2\right)$

$=\frac{10}{8}\left(2t\frac{dt}{dx}\right)$

$=\frac{10}{8}(2t\times \frac{1}{8t^7})$

$=\frac{10}{8}\times \frac{1}{4t^6}$

$=\frac{5}{16t^6}$