If $\sqrt{1-x^2}+\sqrt{1-y^2}=\sqrt{3}(x-y)$, then $d y / d x$ is equal to

$\sqrt{\left(1-x^2\right)\left(1-y^2\right)}$

$\frac{\sqrt{1-x^2}}{\sqrt{1-y^2}}$

$\frac{x^2-y^2}{\sqrt{\left(1-x^2\right)\left(1-y^2\right)}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Analyzing +4 | -1

\sqrt{1-x^2}+\sqrt{1-y^2}=\sqrt{3}(x-y)

, then

d y / d x

is equal to

View answer and explanation

Correct option: B

\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}

Option B (

\frac{{\sqrt{{1-y^2}}}}{{\sqrt{{1-x^2}}}}

) is correct because by substituting x = sin(alpha) and y = sin(beta), and differentiating the resulting expression, we arrive at this result.

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}}