If $ 2 x^{y}+3 y^{x}=20 $, then $ \frac{d y}{d x} $ at $ (2,2) $ is equal to:

$ -\left(\frac{3+\log _{e} 16}{4+\log _{e} 8}\right) $

$ -\left(\frac{3+\log _{e} 8}{2+\log _{e} 4}\right) $

$ -\left(\frac{3+\log _{e} 4}{2+\log _{e} 8}\right) $

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

2 x^{y}+3 y^{x}=20

, then

\frac{d y}{d x}

(2,2)

is equal to:

View answer and explanation

Correct option: A

-\left(\frac{2+\log _{e} 8}{3+\log _{e} 4}\right)

Option A (

-(\frac{{2+\log _{e} 8}}{{3+\log _{e} 4}})

) is correct because it is the result of differentiating the given equation implicitly and then substituting the given point (2,2).

Practice All Chapter MCQs →

a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

C

y

y-e^{x y}+x=0

y = f\left( x \right)

x

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z