If $ (a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2} $, where $ a>b>0 $, then $ \frac{d x}{d y} $ at $ \left(\frac{\pi}{4}, \frac{\pi}{4}\right) $ is:

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}

, where

a>b>0

, then

\frac{d x}{d y}

\left(\frac{\pi}{4}, \frac{\pi}{4}\right)

is:

View answer and explanation

Correct option: C

\frac{a+b}{a-b}

Option C is correct because by differentiating the given equation implicitly and evaluating at the specified point, we find that dx/dy = (a+b)/(a-b).

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}}