let f(x,y)= (xy^2-y)^5 find the partial derivatives of x and y?

Answer 1

∂f(x,y) / ∂x = 5(xy² - y)^4 (y²)

∂f(x,y) / ∂y = 5(xy² - y)^4 (2xy - 1)

Answer 2

Partial derivative w/ respect to x:

5(xy^2-y)^4 * y^2

Partial derivative w/ respect to y:

5(xy^2 - y)^4 * (2xy-1)

The first part is just the chain rule. The second part is the actual partial differentiation. With x, y is considered constant and vice versa.

Answer 3

df/dx = 5(xy^2-y)^4 (y^2)
df/dx = 5(xy^2-y)^4(2xy-1)

Answer 4

Fxy ability first u might desire to locate partial spinoff of z w.r.t x n then multiply with partial spinoff w.r.t y, z=e^2x-3y deldelx(del zdel y) answer wil come out as 0.

Answer 5

df/dx = 5y^6*(xy-1)^4
df/dy = 5y^4*(yx-1)^4*(2yx-1)