Find the first partial derivatives fx and fy of the followin

Find the first partial derivatives f_x and f_y of the following function. f(x, y) = y ln (13x + 13y) f_x= f_y =

Solution

Given that

f( x , y ) = yln( 13x + 13y )

f_{x :}

f_x = / x (  yln( 13x + 13y ) )

f_x = y . / x( ln( 13x + 13y ) )

Apply chain rule , df(u)/dx = df/du.du/dx

Let u = 13x + 13y

/ x( ln( 13x + 13y ) ) = / u( ln(u) ). / x( 13x + 13y )

= 1/u . [ / x(13x) + / x(13y) ] [ since, d/dx( ln x ) = 1/x ]

= 1/u .[ 13 + 0 ]

= 13.( 1/u )

Substitute back u = 13x + 13y

/ x( ln( 13x + 13y ) ) = 13 / ( 13x + 13y )

= 13 / 13( x + y)

= 1 / ( x + y )

Hence,

f_x = y . / x( ln( 13x + 13y ) )

= y.1 / ( x + y )

f_x = y / ( x + y )

Therefore,

f_x =  y / ( x + y )

f_y :

f_y =   / y (  yln( 13x + 13y ) )

Apply the product rule , ( u.v)\' = u\'v + uv\'

u = y , v = ln( 13x + 13y )

f_y =   / y (  yln( 13x + 13y ) )

=   / y (y).ln( 13x + 13y ) + y. / y(  ln( 13x + 13y ) )

= 1. ln( 13x + 13y ) + y [ 1/( 13x +13y ). / y( 13x +13y ) ] [ since,d/dx( ln x ) = 1/x ]

= ln( 13x + 13y ) + y [ 1/( 13x + 13y ).(0 + 13) ]

=  ln( 13x + 13y ) +y [ 13 / 13( x + y ) ]

f_y = ln( 13x +13y ) + ( y / ( x + y ) )

Therefore,

  f_y = ln( 13x +13y ) + ( y / ( x + y ) )