What is (\partial)/(\partial x)(x^2*y^3*(6-x-y)) ?

The answer to (\partial)/(\partial x)(x^2*y^3*(6-x-y)) is y^3(-3x^2-2xy+12x)

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(\partial)/(\partial x)(x^2y^3(6-x-y))

\frac{\partial}{\partial x} (x^{2} \cdot y^{3} \cdot (6 - x - y))

y^{3} (- 3 x^{2} - 2 x y + 12 x)

Solution steps

\frac{\partial}{\partial x} (x^{2} y^{3} (6 - x - y))

Treat

y

as a constant

Take the constant out:

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

= y^{3} \frac{\partial}{\partial x} (x^{2} (6 - x - y))

Apply the Product Rule:

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

f = x^{2}, g = 6 - x - y

= y^{3} (\frac{\partial}{\partial x} (x^{2}) (6 - x - y) + \frac{\partial}{\partial x} (6 - x - y) x^{2})

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (6 - x - y) = - 1

= y^{3} (2 x (6 - x - y) + (- 1) x^{2})

Simplify

y^{3} (2 x (6 - x - y) + (- 1) x^{2}) : y^{3} (- 3 x^{2} - 2 x y + 12 x)

= y^{3} (- 3 x^{2} - 2 x y + 12 x)

n = 3 \sum \infty \frac{3 ^{n}}{n !}

\int 9 cos^{2} (x) d x

\frac{x ^{2} - 3 x}{t ^{2} + 8}

\int \frac{y + 1}{2 y} d y

f (x) = (3 x + 1), at a = 1

What is (\partial)/(\partial x)(x^2*y^3*(6-x-y)) ?
The answer to (\partial)/(\partial x)(x^2*y^3*(6-x-y)) is y^3(-3x^2-2xy+12x)