What is (\partial)/(\partial x)(-3xy(-12-x-y)) ?

The answer to (\partial)/(\partial x)(-3xy(-12-x-y)) is -3y(-2x-y-12)

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(\partial)/(\partial x)(-3xy(-12-x-y))

\frac{\partial}{\partial x} (- 3 x y (- 12 - x - y))

- 3 y (- 2 x - y - 12)

Solution steps

\frac{\partial}{\partial x} (- 3 x y (- 12 - x - y))

Treat

y

as a constant

Take the constant out:

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

= - 3 y \frac{\partial}{\partial x} (x (- 12 - x - y))

Apply the Product Rule:

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

f = x, g = - 12 - x - y

= - 3 y (\frac{\partial}{\partial x} (x) (- 12 - x - y) + \frac{\partial}{\partial x} (- 12 - x - y) x)

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

\frac{\partial}{\partial x} (- 12 - x - y) = - 1

= - 3 y (1 \cdot (- 12 - x - y) + (- 1) x)

Simplify

- 3 y (1 \cdot (- 12 - x - y) + (- 1) x) : - 3 y (- 2 x - y - 12)

= - 3 y (- 2 x - y - 12)

\frac{d}{d x} (- \frac{2}{3} x^{3})

\int_{- 1}^{4} (3 x^{2} + 8 x - 5) d x

\int 2 x^{4} + e^{2 x} + sin (3 x) d x

f (x) = \frac{4}{1 + x ^{2}}

\int x^{2} \cdot e^{2 x} d x

What is (\partial)/(\partial x)(-3xy(-12-x-y)) ?
The answer to (\partial)/(\partial x)(-3xy(-12-x-y)) is -3y(-2x-y-12)