For $\left| x \right| < 1,$ let $y = 1 + x + {x^2} + \ldots $ to $\infty ,$ then $\frac{{dy}}{{dx}} - y$ is equal to

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

For

\left| x \right| < 1,

let

y = 1 + x + {x^2} + \ldots

\infty ,

then

\frac{{dy}}{{dx}} - y

is equal to

View answer and explanation

Correct option: D

x{y^2}

Option D (

x{y^2}

) is correct because, after finding the derivative

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

and simplifying, the expression

\frac{{dy}}{{dx}} - y

is equal to

x{y^2}

Practice All Chapter MCQs →

f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

x>0

x y=1

x+y

f(x)=[x] \sin (\pi x)

[x]

x

x=k, k

f(x)

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

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

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}