If $2^x+2^y=2^{x+y},$ then $\frac{d y}{d x}$ is equal to

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

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

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

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

2^x+2^y=2^{x+y},

then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is derived by differentiating the given equation implicitly and solving for

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

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}}