JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

x = \frac{{2at}}{{1 + {t^3}}}

and

y = \frac{{2a{t^2}}}{{{{\left( {1 + {t^3}} \right)}^2}}}

, then

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

View answer and explanation

Correct option: C

\frac{x}{a}

The correct answer is C (

\frac{x}{a}

) because the relationship between x and y allows for a direct substitution and differentiation.

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