What is (\partial)/(\partial y)(4x^2+2xy+y^2-8) ?

The answer to (\partial)/(\partial y)(4x^2+2xy+y^2-8) is 2x+2y

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(\partial)/(\partial y)(4x^2+2xy+y^2-8)

\frac{\partial}{\partial y} (4 x^{2} + 2 x y + y^{2} - 8)

2 x + 2 y

Solution steps

\frac{\partial}{\partial y} (4 x^{2} + 2 x y + y^{2} - 8)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (4 x^{2}) + \frac{\partial}{\partial y} (2 x y) + \frac{\partial}{\partial y} (y^{2}) - \frac{\partial}{\partial y} (8)

\frac{\partial}{\partial y} (4 x^{2}) = 0

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

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

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

= 0 + 2 x + 2 y - 0

Simplify

= 2 x + 2 y

\frac{d}{d x} (x^{5} e^{x y} + 3 x)

f (x) = tan (5 x)

∣ 3 x ∣, x^{2} - 4

ln (\frac{( x - 3 )}{x - 1})

\int_{1}^{4} \frac{4}{x} d x

What is (\partial)/(\partial y)(4x^2+2xy+y^2-8) ?
The answer to (\partial)/(\partial y)(4x^2+2xy+y^2-8) is 2x+2y