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

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

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

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

2 y - x

Solution steps

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

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

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

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

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

= 0 + 2 y - x

Simplify

= 2 y - x

x e^{x - 1}

\int e^{7 x} d x

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

f (x) = 13 x

f (x) = 10 x, at x = 10

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