JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

\sqrt {1 - {x^6}} + \sqrt {1 - {y^6}} = a\left( {{x^3} - {y^3}} \right)

and

\frac{{dy}}{{dx}} = f\left( {x,y} \right)\sqrt {\frac{{1 - {y^6}}}{{1 - {x^6}}}}

, then

f\left( {x,y} \right) =

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because by substituting

{x^3} = \sin A

and

{y^3} = \sin B

, differentiating, and simplifying, we find that

f(x,y) = \frac{{{x^2}}}{{{y^2}}}

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