JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

x=\sin ^{-1} t

and

y=\log \left(1-t^2\right)

; then

\left.\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right|_{t=1 / 2}

View answer and explanation

Correct option: A

-8 / 3

Option A (-8/3) is correct because it is the value of the second derivative of

y

with respect to

x

evaluated at

t = 1/2

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