JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

y = x + {x^2} + {x^3} \ldots \infty ,

where

\left| x \right| < 1,

then for

\left| y \right| < 1,\frac{{dx}}{{dy}}

is equal to

View answer and explanation

Correct option: C

1 - 2y + 3{y^2} - ...\infty

The correct answer is C (

1 - 2y + 3{y^2} - ...\infty

) because it is derived by expressing x in terms of y and then differentiating the resulting series.

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