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

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

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

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

4 y - 2 x

Solution steps

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

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

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

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

= 0 - 2 x + 4 y

Simplify

= 4 y - 2 x

f (x) = x^{3} - 4 x, (2, 0)

cos (x) - \frac{2}{2} x

x \to 1^{- 1} lim (- 1.6 x - 0.6)

\frac{d}{d x} (cot (3 x^{2} + 5))

π (2 t - 3)

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