Question

If y = t¹⁰ + 1 and x = t⁸ + 1, then $\frac{d^{2}y}{d{x}^2}$ = ___________________.

Answer 1

Given, y = t¹⁰ + 1 and x = t⁸ + 1.

$y=t^{10}+1$

Differentiating both sides with respect to t, we get

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

$x=t^8+1$

Differentiating both sides with respect to t, we get

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

$$\begin{gathered} \therefore \frac{dy}{dx}=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}} \\ \Rightarrow \frac{dy}{dx}=\frac{10t^9}{8t^7} \\ \Rightarrow \frac{dy}{dx}=\frac{5t^2}{4} \end{gathered}$$

Differentiating both sides with respect to x, we get

$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{5t^2}{4}\right)$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{5}{4}\times 2t\frac{dt}{dx}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{5}{2}t\times \frac{1}{8t^7}\left(\frac{dx}{dt}=8t^7\Rightarrow \frac{dt}{dx}=\frac{1}{8t^7}\right)$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{5}{16t^6}$


If y = t¹⁰ + 1 and x = t⁸ + 1, then $\frac{d^{2}y}{d{x}^2}$ = $\underline{\frac{5}{16t^6}}$.