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

$\frac{{{2^x} + {2^y}}}{{{2^x} - {2^y}}}$

$\frac{{{2^x} + {2^y}}}{{1 + {2^{x + y}}}}$

$\frac{{{2^{x + y}} - {2^x}}}{{{2^y}}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Analyzing +4 | -1

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

, then

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the result of implicit differentiation and algebraic manipulation of the given equation.

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}}