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

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

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

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

2 x - 2

Solution steps

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

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} (x^{2}) = 2 x

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

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

= 2 x + 0 - 2

Simplify

= 2 x - 2

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

\int 16 cos^{2} (x) d x

y^{^{''}} - 4 y = \frac{( e ^{2 x} )}{( x )}

x \to - 3 lim (- x^{2} - x + 4)

x \to 0 lim (x 2)

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