What is (\partial)/(\partial y)(8xy+8y^2-2x+2y-1) ?

The answer to (\partial)/(\partial y)(8xy+8y^2-2x+2y-1) is 8x+16y+2

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(\partial)/(\partial y)(8xy+8y^2-2x+2y-1)

\frac{\partial}{\partial y} (8 x y + 8 y^{2} - 2 x + 2 y - 1)

8 x + 16 y + 2

Solution steps

\frac{\partial}{\partial y} (8 x y + 8 y^{2} - 2 x + 2 y - 1)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (8 x y) + \frac{\partial}{\partial y} (8 y^{2}) - \frac{\partial}{\partial y} (2 x) + \frac{\partial}{\partial y} (2 y) - \frac{\partial}{\partial y} (1)

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

\frac{\partial}{\partial y} (8 y^{2}) = 16 y

\frac{\partial}{\partial y} (2 x) = 0

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

\frac{\partial}{\partial y} (1) = 0

= 8 x + 16 y - 0 + 2 - 0

Simplify

= 8 x + 16 y + 2

\frac{d}{d x} ((1 + \frac{5}{x})^{x})

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

\int_{0}^{\frac{2}{3}} \frac{1}{4 + 9 x ^{2}} d x

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What is (\partial)/(\partial y)(8xy+8y^2-2x+2y-1) ?
The answer to (\partial)/(\partial y)(8xy+8y^2-2x+2y-1) is 8x+16y+2