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

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

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

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

2 y - x

Solution steps

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

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

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

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

= 0 - x + 2 y

Simplify

= 2 y - x

\int \frac{x}{( x + 2 ) ( x - 4 )} d x

\int sec (2 x) \cdot tan (2 x) d x

\int \frac{2 x - 3}{x ^{2} + 4} d x

n = 1 \sum \infty \frac{2 n}{n ^{2}}

f (x) = 7 arctan (7 x)

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