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

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

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

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

2 y - 2 x

Solution steps

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

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

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

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

= 0 - 2 x + 2 y

Simplify

= 2 y - 2 x

y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = - 2

\int x x^{2} + a^{2} d x

\int \frac{( 2 + x ^{2} )}{1 + x ^{2}} d x

\frac{\partial}{\partial x} (ln (x + 5 y + 6 z))

\frac{d}{d x} (x sec (x))

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