Let $ y=y(x) $ be $ a $ function of $ x $ satisfying $ y \sqrt{1-x^{2}}=k-x \sqrt{1-y^{2}} $ where $ k $ is $ a $ constant and $ y\left(\frac{1}{2}\right)=-\frac{1}{4} $. Then $ \frac{d y}{d x} $ at $ x=\frac{1}{2} $, is equal to:

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

Let

y=y(x)

a

function of

x

satisfying

y \sqrt{1-x^{2}}=k-x \sqrt{1-y^{2}}

where

k

a

constant and

y\left(\frac{1}{2}\right)=-\frac{1}{4}

. Then

\frac{d y}{d x}

x=\frac{1}{2}

, is equal to:

View answer and explanation

Correct option: B

-\frac{\sqrt{5}}{2}

Option B (

-\frac{\sqrt{5}}{2}

) is correct because it is the value of

\frac{dy}{dx}

obtained by implicit differentiation and substitution of the given values.

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}}