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

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

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

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

2 y - 4

Solution steps

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

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 y)

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

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

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

= 0 + 2 y - 4

Simplify

= 2 y - 4

\int x arctan (16 x) d x

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

\frac{d y}{d x} = 4 e^{- 3 x}

\int e^{x} \cdot sin (e^{x}) d x

x \to 3 - lim (- (3)^{2} + 5 (3) - 5)

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