What is (\partial)/(\partial y)((x-y)(1-xy)) ?

The answer to (\partial)/(\partial y)((x-y)(1-xy)) is -x^2+2xy-1

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(\partial)/(\partial y)((x-y)(1-xy))

\frac{\partial}{\partial y} ((x - y) (1 - x y))

- x^{2} + 2 x y - 1

Solution steps

\frac{\partial}{\partial y} ((x - y) (1 - x y))

Treat

x

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x - y, g = 1 - x y

= \frac{\partial}{\partial y} (x - y) (1 - x y) + \frac{\partial}{\partial y} (1 - x y) (x - y)

\frac{\partial}{\partial y} (x - y) = - 1

\frac{\partial}{\partial y} (1 - x y) = - x

= (- 1) (1 - x y) + (- x) (x - y)

Simplify

(- 1) (1 - x y) + (- x) (x - y) : - x^{2} + 2 x y - 1

= - x^{2} + 2 x y - 1

x y^{^{'}} + 2 y = \frac{4 e ^{3 x}}{x}

f (x) = 2 x^{3} - 21 x^{2} + 72 x - 5

\frac{\partial}{\partial x} (e^{2 x - y})

x \to 0 lim (\frac{x - 14}{x - 2})

x \to 0 + lim (\frac{1}{x ^{2}})

What is (\partial)/(\partial y)((x-y)(1-xy)) ?
The answer to (\partial)/(\partial y)((x-y)(1-xy)) is -x^2+2xy-1