JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

Let a function

y=y(x)

be defined parametrically by

x=2 t-|t|, y=t^2+t|t|

. Then

y^{\prime}(x), x>0

View answer and explanation

Correct option: B

4 x

Option C (

4x

) is correct because after expressing

y

in terms of

x

for

x>0

, differentiating yields

y'(x) = 4x

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x

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[x]

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\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

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