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

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

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

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

2 x - y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

= 2 x - y + 0

Simplify

= 2 x - y

\frac{d}{d x} (sin (e^{x^{2}}))

\int_{1}^{e} ln (21 x) d x

x \to - 5 lim (x + 5)

\frac{\partial}{\partial z} (y^{2} z e^{z})

\frac{\partial}{\partial x} ((x^{2} + y^{2} + z^{2})^{- 0.5})

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