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

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

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

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

2 x - 2

Solution steps

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

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

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

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

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

= 2 x + 0 - 2 - 0

Simplify

= 2 x - 2

\frac{d}{d x} (\frac{x ^{2} - 4}{x ^{2}})

f (x) = e^{4 x}, at x = \frac{1}{4} ln (5)

f (x) = 2 sin (x) + sin (2 x), y = 0, 0 \leq x \leq π

x^{^{'}} = x - 15

\frac{d}{d x} (\frac{2}{( 2 + x ) ^{3}})

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