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

The answer to (\partial)/(\partial x)(x^3y^2(6-x-y)) is y^2(-4x^3-3x^2y+18x^2)

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

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

y^{2} (- 4 x^{3} - 3 x^{2} y + 18 x^{2})

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{3}, g = 6 - x - y

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

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

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

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

Simplify

y^{2} (3 x^{2} (6 - x - y) + (- 1) x^{3}) : y^{2} (- 4 x^{3} - 3 x^{2} y + 18 x^{2})

= y^{2} (- 4 x^{3} - 3 x^{2} y + 18 x^{2})

\frac{d}{d x} (\frac{x - 3}{x + 5})

\int sin^{2} (\frac{1}{3} x) d x

\frac{s}{( s ^{2} + 5 ^{2} ) ^{2}}

\int \frac{4 + x ^{2}}{x} d x

\frac{d}{d x} (\frac{x}{( x - 1 )})

What is (\partial)/(\partial x)(x^3y^2(6-x-y)) ?
The answer to (\partial)/(\partial x)(x^3y^2(6-x-y)) is y^2(-4x^3-3x^2y+18x^2)