Let y=t10+1 and x=t8+1, then d2ydx2 is equal to.

y=t10+1,x=t8+1dydt=10t9,dxdt=8t7⇒dydx=dy/dtdx/dt=10t98t7=54t2⇒d2ydx=54(2t)dtdx=54×2t×18t7=516t6

Let y=t10+1 and x=t8+1, then d2ydx2 is equal to.

y=t10+1,x=t8+1dydt=10t9,dxdt=8t7⇒dydx=dy/dtdx/dt=10t98t7=54t2⇒d2ydx=54(2t)dtdx=54×2t×18t7=516t6

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

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None of these

$\frac{5}{2} t$

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

$20 t^{8}$

answer is C.

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Detailed Solution

$\begin{array}{l} y = t^{10} + 1, x = t^{8} + 1 \\ \frac{d y}{d t} = 10 t^{9}, \frac{d x}{d t} = 8 t^{7} \\ \Rightarrow \frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{10 t^{9}}{8 t^{7}} = \frac{5}{4} t^{2} \\ \Rightarrow \frac{d^{2} y}{d x} = \frac{5}{4} (2 t) \frac{d t}{d x} \\ = \frac{5}{4} \times 2 t \times \frac{1}{8 t^{7}} \\ = \frac{5}{16 t^{6}} \end{array}$