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

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

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

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

2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

= 0 + 2 y - 0

Simplify

= 2 y

\int arcsin (x) d x

\frac{d}{d x} (2 x^{3} - 6 x^{2} - 18 x - 24)

\frac{d y}{d x} = \frac{y ^{2} + x x ^{2} + y ^{2}}{x y}

\int (x + 1) e^{2 x^{2} + 4 x} d x

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