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

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

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

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

2 x + 1 - 2 y

Solution steps

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

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

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

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

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

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

= 2 x - 2 y + 0 + 1 - 0

Simplify

= 2 x + 1 - 2 y

\frac{d}{d x} (cos (π \cdot x))

x^{2} \cdot e^{x} + 4 \cdot y \cdot e^{(- y)} \cdot y^{^{'}} = 0

\frac{d}{d x} ((x - 1) ln (x - 1))

\int_{0}^{x^{2}} t^{2} d t

y = 5 x^{4} - x^{2} + 6

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