What is (\partial)/(\partial x)(8y-(x+y)^3) ?

The answer to (\partial)/(\partial x)(8y-(x+y)^3) is -3y^2-6yx-3x^2

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(\partial)/(\partial x)(8y-(x+y)^3)

\frac{\partial}{\partial x} (8 y - (x + y)^{3})

- 3 y^{2} - 6 y x - 3 x^{2}

Solution steps

\frac{\partial}{\partial x} (8 y - (x + y)^{3})

Simplify

8 y - (x + y)^{3} : 8 y - x^{3} - 3 y x^{2} - 3 y^{2} x - y^{3}

= \frac{\partial}{\partial x} (8 y - x^{3} - 3 y x^{2} - 3 y^{2} x - y^{3})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (8 y) - \frac{\partial}{\partial x} (x^{3}) - \frac{\partial}{\partial x} (3 y x^{2}) - \frac{\partial}{\partial x} (3 y^{2} x) - \frac{\partial}{\partial x} (y^{3})

\frac{\partial}{\partial x} (8 y) = 0

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

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

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

\frac{\partial}{\partial x} (y^{3}) = 0

= 0 - 3 x^{2} - 6 y x - 3 y^{2} - 0

Simplify

= - 3 y^{2} - 6 y x - 3 x^{2}

\frac{d}{d x} (5 cos^{2} (π x))

\frac{d}{d x} (\frac{2 x + 1}{1 - x})

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

\int \frac{1}{4 x ^{2} - 1} d x

\frac{d}{d x} (2 x^{7} - 3 x^{4} + 4 x^{3} - 12 x + 4)

What is (\partial)/(\partial x)(8y-(x+y)^3) ?
The answer to (\partial)/(\partial x)(8y-(x+y)^3) is -3y^2-6yx-3x^2