What is (\partial)/(\partial y)(x^2-2xy+3y^2) ?

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

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(\partial)/(\partial y)(x^2-2xy+3y^2)

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

6 y - 2 x

Solution steps

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

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

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

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

= 0 - 2 x + 6 y

Simplify

= 6 y - 2 x

\int sin^{3} (2 x) cos^{6} (2 x) d x

x = 4 y, x = 0, y = 4

\frac{e ^{x}}{2 x ^{2}}

\frac{d}{d x} (\frac{x}{x - 1})

(x^{5} + 2 x) \cdot (x^{3} + x^{2} + x + 7)

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