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

The answer to (\partial)/(\partial y)(x^2-xy+y^2-y) is 2y-x-1

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

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

2 y - x - 1

Solution steps

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

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

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

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

\frac{\partial}{\partial y} (y) = 1

= 0 - x + 2 y - 1

Simplify

= 2 y - x - 1

\int x^{5} (cos (x^{3})) d x

\frac{d}{d z} (ln (\frac{a ^{2} - z ^{2}}{a ^{2} + z ^{2}}))

\frac{d}{d x} (\frac{1}{3} π r (x) (x)^{2} h)

\int_{0}^{7} \frac{1}{9 + x ^{2}} d x

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

What is (\partial)/(\partial y)(x^2-xy+y^2-y) ?
The answer to (\partial)/(\partial y)(x^2-xy+y^2-y) is 2y-x-1