Let y = t^{10} + 1 and x = t^8 + 1, then frac | vedclass

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(C) Given $y = t^{10} + 1$ and $x = t^8 + 1.$First,we express $t$ in terms of $x$: $t^8 = x - 1 \Rightarrow t^2 = (x - 1)^{1/4}.$Substitute $t^2$ into

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

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(C) Given $y = t^{10} + 1$ and $x = t^8 + 1.$First,we express $t$ in terms of $x$: $t^8 = x - 1 \Rightarrow t^2 = (x - 1)^{1/4}.$Substitute $t^2$ into the expression for $y$: $y = (t^2)^5 + 1 = ((x - 1)^{1/4})^5 + 1 = (x - 1)^{5/4} + 1.$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{5}{4}(x - 1)^{5/4 - 1} = \frac{5}{4}(x - 1)^{1/4}.$Differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = \frac{5}{4} 	imes \frac{1}{4}(x - 1)^{1/4 - 1} = \frac{5}{16}(x - 1)^{-3/4} = \frac{5}{16(x - 1)^{3/4}}.$Since $x - 1 = t^8$,we have $(x - 1)^{3/4} = (t^8)^{3/4} = t^6.$Therefore,$\frac{d^2y}{dx^2} = \frac{5}{16t^6}.$

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

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