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

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

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

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

2 x - 2 y

Solution steps

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

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

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

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

= 2 x + 0 - 2 y

Simplify

= 2 x - 2 y

t \to 1 lim (\frac{ln ( t )}{t - 1})

y^{^{'}} + x y = 4 x y^{\frac{7}{2}}

\int_{0}^{\frac{π}{2}} 4 cos^{2} (x) d x

\frac{d}{d x} ((- 9 x^{2} + 25)^{2})

\int \frac{1}{x ( ln ( x ) ) ^{7}} d x

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