What is (\partial)/(\partial x)(x^2+y^2-4y) ?

The answer to (\partial)/(\partial x)(x^2+y^2-4y) is 2x

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(\partial)/(\partial x)(x^2+y^2-4y)

\frac{\partial}{\partial x} (x^{2} + y^{2} - 4 y)

2 x

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} - 4 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} (4 y)

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

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

\frac{\partial}{\partial x} (4 y) = 0

= 2 x + 0 - 0

Simplify

= 2 x

What is (\partial)/(\partial x)(x^2+y^2-4y) ?
The answer to (\partial)/(\partial x)(x^2+y^2-4y) is 2x