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

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

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(\partial)/(\partial x)((x-1)^2+xy+(y-1)^2)

\frac{\partial}{\partial x} ((x - 1)^{2} + x y + (y - 1)^{2})

2 x + y - 2

Solution steps

\frac{\partial}{\partial x} ((x - 1)^{2} + x y + (y - 1)^{2})

Simplify

(x - 1)^{2} + x y + (y - 1)^{2} : x^{2} + x y - 2 x + y^{2} + 2 - 2 y

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

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

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

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

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

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

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

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

Simplify

= 2 x + y - 2

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