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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = x, g = - 3 - x - y

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

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

\frac{\partial}{\partial x} (- 3 - x - y) = - 1

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

Simplify

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

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

\frac{d y}{d x} = \frac{( x ^{2} + y )}{x ^{3}}

f (x) = ln (x + 2)

\int_{8}^{9} \frac{1}{x ^{2} - 1} d x

y = 9 - x^{2}, 0 \leq x \leq 2

f (x) = 7 e^{x^{2} - 9 x + 8}, at x = 0

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