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

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

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

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

2 x - 2 y

Solution steps

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

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

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

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

= 2 x - 2 y + 0

Simplify

= 2 x - 2 y

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

\int \frac{y + 1}{y - 1} d y

y^{^{''}} - 5 y^{^{'}} = x - 2

3 x^{2} cos (x) cot (x)

e^{x} sin (2 x)

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