JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

x=\cos \theta, y=\sin ^3 \theta

, then

\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}

\theta=\pi / 2

View answer and explanation

Correct option: D

-3

Option D (-3) is correct because after finding the first and second derivatives and substituting

\theta = \frac{{\pi}}{{2}}

, the expression evaluates to -3.

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