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

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

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

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

2 x + y + 2

Solution steps

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

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

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

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

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

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

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

= 2 x + y + 0 + 2 - 0

Simplify

= 2 x + y + 2

\frac{d}{d t} (- cos (2 t))

y = x^{8} e^{- x}, (1, \frac{1}{e})

\int (2 x^{4} - 8 x^{3}) \cdot e^{- 3 x} d x

\frac{d}{d x} (\frac{2 cos ( x )}{sec ( x )})

y^{^{''}} + y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 0

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