What is (\partial)/(\partial y)(x(1-x-2y)) ?

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

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(\partial)/(\partial y)(x(1-x-2y))

\frac{\partial}{\partial y} (x (1 - x - 2 y))

- 2 x

Solution steps

\frac{\partial}{\partial y} (x (1 - x - 2 y))

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= x \frac{\partial}{\partial y} (1 - x - 2 y)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= x (\frac{\partial}{\partial y} (1) - \frac{\partial}{\partial y} (x) - \frac{\partial}{\partial y} (2 y))

\frac{\partial}{\partial y} (1) = 0

\frac{\partial}{\partial y} (x) = 0

\frac{\partial}{\partial y} (2 y) = 2

= x (0 - 0 - 2)

Simplify

= - 2 x

(0, 8), (8, 1)

x, at x = 16

\int_{2}^{x} (3 t^{2} - 1) d t

F (r) = \frac{5}{r ^{3}}

\frac{d}{d x} (\frac{3}{x + 7})

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