Question

find dy/dx of x log y+ y log x=0​

Answer 1

Answer:

- y ( y + x logy ) / x ( x + ylogx )

Step-by-step explanation:

Given---> x logy + ylogx = 0

To find---> Derivative of given equation

Solution---> ATQ,

x logy + y logx = 0

=> x logy = - y logx

Differrntiating with respect to x, we get,

=> d / dx ( x logy ) = - d / dx ( y logx )

=> x d / dx ( logy ) + logy d / dx ( x )

= - { y d / dx ( logx ) + ( logx ) d / dx ( y ) }

=> x ( 1 / y ) dy/dx + logy ( 1 ) =

- { y ( 1 / x ) + logx dy / dx }

=> ( x / y ) dy / dx + logy = - y/x - logx dy / dx

=> ( x / y ) dy / dx + logx dy / dx = - logy - y / x

=> { ( x / y ) + logx } dy / dx = - ( x logy + y ) / x

=> {( x + y logx ) / y } dy/dx = - ( x logy + y ) / x

=> dy /d = - y ( x logy + y ) / x ( x + y logx )

Formulee applied--->

1) d/dx ( u v )= u dv/dx + v du/dx

2) d/dx ( x ) = 1

3) d/dx ( logx ) = 1 / x

Answer 2

Answer:

sorry don't know the answer XD