Question #5412

I need to find the partial derivatives of the following functions
1.) f(x,y)=ln((x-y)/(x+y)^2)
2.) f(x,y)=ln(sqrt(xy))

Expert's answer

Let's find the partial derivatives of the following functions

1.) f(x,y)=ln((x-y)/(x+y)²)

df(x,y)/dx =

& = (((x-y)/(x+y)²))' / ( (x-y)/(x+y)² ) =

& = ( (x-y)'(x+y)² - ((x+y)²)'(x-y) ) / ( (x+y)^4(x-y)/(x+y)² ) =

& = ( (x+y)² - 2(x+y)(x-y) ) / ( (x+y)²(x-y) ) =

& = ( (x+y) - 2(x-y) ) / ( (x+y)(x-y) ) =

& = (3y-x)/(x²-y²).

df(x,y)/dy = (y-3x)/(x²-y²).

2.) f(x,y)=ln(sqrt(xy))

df(x,y)/dx =

& = (sqrt(xy))' / (sqrt(xy)) =

& = 0.5(xy)^(-0.5)*y / sqrt(xy) =

& = 0.5y / (xy) = 0.5/x = 1/(2x)

df(x,y)/dy = 1/(2y)

1.) f(x,y)=ln((x-y)/(x+y)²)

df(x,y)/dx =

& = (((x-y)/(x+y)²))' / ( (x-y)/(x+y)² ) =

& = ( (x-y)'(x+y)² - ((x+y)²)'(x-y) ) / ( (x+y)^4(x-y)/(x+y)² ) =

& = ( (x+y)² - 2(x+y)(x-y) ) / ( (x+y)²(x-y) ) =

& = ( (x+y) - 2(x-y) ) / ( (x+y)(x-y) ) =

& = (3y-x)/(x²-y²).

df(x,y)/dy = (y-3x)/(x²-y²).

2.) f(x,y)=ln(sqrt(xy))

df(x,y)/dx =

& = (sqrt(xy))' / (sqrt(xy)) =

& = 0.5(xy)^(-0.5)*y / sqrt(xy) =

& = 0.5y / (xy) = 0.5/x = 1/(2x)

df(x,y)/dy = 1/(2y)

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