Let y = t^10 + 1 and x = t^8 + 1 , then d^2y/dx^2 is: - Mathematics MCQ Mathematics Mock Test - 6 2026 | ExamTest.live

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

Solution & Step-by-step Explanation

Given y = t^{10} + 1 and x = t^{8} + 1 .Differentiating with respect to t : \frac{d y}{d t} = 10 t^{9} \frac{d x}{d t} = 8 t^{7} Finding \frac{d y}{d x} : \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} Finding \frac{d ^{2} y}{d x ^{2}} : \frac{d ^{2} y}{d x ^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d t} (\frac{5}{4} t^{2}) \cdot \frac{d t}{d x} \frac{d ^{2} y}{d x ^{2}} = (\frac{5}{4} \cdot 2 t) \cdot \frac{1}{8 t ^{7}} = \frac{5 t}{2} \cdot \frac{1}{8 t ^{7}} \frac{d ^{2} y}{d x ^{2}} = \frac{5}{16 t ^{6}}

Try it yourself before checking the explanation above.

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

\frac{5}{2} t

20 t^{8}

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

None of these

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