JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

{y^2} = a{x^2} + bx + c,

where

a,b,c

are constants,then

{y^3}\frac{{{d^2}y}}{{d{x^2}}}

is equal to

View answer and explanation

Correct option: A

a constant

Option A (a constant) is correct because after differentiating twice and simplifying,

{y^3}\frac{{{d^2}y}}{{d{x^2}}}

equals a constant value of

\frac{{4ac - {b^2}}}{4}

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