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 the given equation twice and simplifying, the expression

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

results in a constant value.

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