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

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

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

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

- 3 y^{2} - 6 y x + 8 - 3 x^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

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

= 8 - 0 - 3 x^{2} - 6 x y - 3 y^{2}

Simplify

= - 3 y^{2} - 6 y x + 8 - 3 x^{2}

2 sin^{4} (x)

\int \frac{1}{( 2 - y ) ( 1 - y )} d y

\frac{\partial}{\partial x} (2 ln (x) + 5 ln (y))

\int (5 x^{3} + 6 x) e^{x^{2}} d x

\int \frac{1}{9 y ^{2} + 4} d y

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