Let y = t¹⁰ + 1 and x = t⁸ + 1, then d²y/dx², is equal to

The correct option is "5/16t⁶".

Let us solve this step-by-step using parametric differentiation.

We are given the parametric equations:
$y=t^{10}+1$
and
$x=t^8+1$

First, we find the first derivative of both $y$ and $x$ with respect to the parameter $t$:
$\frac{dy}{dt}=\frac{d}{dt}(t^{10}+1)=10t^9$
$\frac{dx}{dt}=\frac{d}{dt}(t^8+1)=8t^7$

Next, we find the first derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$, using the chain rule:
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

Substituting the values we calculated:
$\frac{dy}{dx}=\frac{10t^9}{8t^7}=\frac{5}{4}{t}^2$

Now, we need to find the second derivative, $\frac{d^{2}y}{d{x}^2}$. By applying the chain rule again, we have:
$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)·\frac{dt}{dx}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$

First, differentiate $\frac{dy}{dx}$ with respect to $t$:
$\frac{d}{dt}\left(\frac{5}{4}{t}^2\right)=\frac{5}{4}·2t=\frac{5}{2}t$

Now, substitute this and $\frac{dx}{dt}=8t^7$ into the expression for the second derivative:
$\frac{d^{2}y}{d{x}^2}=\frac{\frac{5}{2}t}{8t^7}=\frac{5}{2·8}·\frac{t}{t^7}$

Simplifying the fraction gives:
$\frac{d^{2}y}{d{x}^2}=\frac{5}{16}·\frac{1}{t^6}=\frac{5}{16t^6}$

This matches the correct option "5/16t⁶" (where the term represents $\frac{5}{16t^6}$).