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

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

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

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

8 y - 4 x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (4 x y) = 4 x

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

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

= 0 - 4 x + 8 y + 0

Simplify

= 8 y - 4 x

n = 1 \sum \infty cos (\frac{2}{n})

x \to 1 + lim (\frac{7}{x ^{3} - 1})

x \frac{d y}{d x} + 3 y = 2 x^{5}, y (2) = 7

(83) (8 - 7)

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

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