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

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

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

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

2 x

Solution steps

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

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 y) + \frac{\partial}{\partial x} (1)

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

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

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

\frac{\partial}{\partial x} (1) = 0

= 2 x + 0 - 0 + 0

Simplify

= 2 x

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

\int_{1}^{2} \frac{1}{x ( ln ( x ) ) ^{p}} d x

f (x) = ln (3 x e^{- x})

\frac{f ( x )}{x ^{5}}

\frac{\partial}{\partial x} (sin (x^{2} + y^{3}))

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