JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

Let

y = {t^{10}} + 1

and

x = {t^8} + 1,

then

\frac{{{d^2}y}}{{d{x^2}}}

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because after finding the first and second derivatives using parametric differentiation, the expression simplifies to this result.

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