Find the partial derivatives with respect to both x and y: f(x,y) = ln ((x-y)/(x+y)^2) Zx = ln(x-y)-ln(x+y)^2 = lnx-ln2x = 1/x - 2/x or Zx = ln(x-y)-ln(x+y)^2 = 1/(x-y) - 2/(x+y)^2 Zy =...
Find the partial derivatives with respect to both x and y: f(x,y) = ln ((x-y)/(x+y)^2)
Zx = ln(x-y)-ln(x+y)^2 = lnx-ln2x = 1/x - 2/x
or Zx = ln(x-y)-ln(x+y)^2 = 1/(x-y) - 2/(x+y)^2
Zy = ln(x-y)-ln(x+y)^2 = lny - 2lny = 1/y - 2/y
or Zy = ln(x-y)-ln(x+y)^2 = 1/(x-y) - 2/(x+y)^2
not sure which is the right way to do it, and if it can be rewritten any m ore useing the properties of logarithms.
Thank you for your help.
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`f(x,y) = ln ((x-y)/(x+y)^2)`
`f(x,y) = ln(x-y)-ln(x+y)^2`
`f(x,y) = ln(x-y)-2ln(x+y)`
When we are doing partial derivatives we consider only one changing variable and others are constant.
When you derivate with respect to x we consider y as a constant and derivation with respect to y, x is considered as constant.
`(delf(x,y))/(delx) = f_x(x,y)`
`(delf(x,y))/(dely) = f_y(x,y)`
`f_x(x,y) = 1/(x-y)-2/(x+y)`
`f_y(x,y) = 1/(x-y)*(-1)-2/(x+y) = (-1)[1/(x-y)+2/(x+y)]`
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