Partial Differentation - Please help!?

z=x²y²/√(x+y)

To partially differentiate w.r.t. x, treat y as a constant, and vice versa

Use the chain rule. u=x²y² , v=1/√(x+y)

∂z/∂x = 2xy²/√(x+y) - ½x²y²/√(x+y)³

Similarly for ∂z/∂x but with roles reversed.

∂z/∂y = 2x²y/√(x+y) - ½x²y²/√(x+y)³

When doing partial differentiation let`s say for dz/dx you consider y as a constant and just x the variable

dz/dx = {( 2y²x )*sqrt(x+y) - x²y²/[2sqrt(x+y)]}/(x+y)

for understanding better just substitute y with a number and differentiate normally... until you get used to it.(but of course insteand of writing the number you write "y"

and when doing dz/dy x becomes a constant and y the variable

dz/dy = [(2x²y)*sqrt(x+y) - (x²y²)/2sqrt(x+y)]/(x+1)

To find ∂z/∂x, y should be treated as a constant and differentiate z w.r.t. x. Thus,

ln z = 2lnx + 2lny - (1/2)ln(x + y)

=> (1/z) ∂z/∂x = 2/x - (1/2) * 1/(x+y)

=> ∂z/∂x = (x^2y^2)/sqrt(x+y) * [2/x - (1/2) * 1/(x+y)]

Similarly,

∂z/∂y = (x^2y^2)/sqrt(x+y) * [2/y - (1/2) * 1/(x+y)]

this question is a sprint problematical using reality that we've been given to apply quotent rule. besides here is going. you're good say that as quickly as looking dx we enable y be consistent. for this reason dz/dx = [ln(xy) * y - xy * (y / xy) ] / [ (ln(xy))^2] dz/dy = [ln(xy) * x - xy * (x / xy) ] / [ (ln(xy))^2]