What is (\partial)/(\partial y)(x^3+x^2y-y^2-4y) ?

The answer to (\partial)/(\partial y)(x^3+x^2y-y^2-4y) is x^2-2y-4

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(\partial)/(\partial y)(x^3+x^2y-y^2-4y)

\frac{\partial}{\partial y} (x^{3} + x^{2} y - y^{2} - 4 y)

x^{2} - 2 y - 4

Solution steps

\frac{\partial}{\partial y} (x^{3} + x^{2} y - 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^{3}) + \frac{\partial}{\partial y} (x^{2} y) - \frac{\partial}{\partial y} (y^{2}) - \frac{\partial}{\partial y} (4 y)

\frac{\partial}{\partial y} (x^{3}) = 0

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

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

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

= 0 + x^{2} - 2 y - 4

Simplify

= x^{2} - 2 y - 4

\int_{- 1}^{2} (x - 1) (2 x + 3) d x

\frac{d}{d x} (- 64 e^{- 4 x})

\int y cos (x) y d x

x \to 0 lim (\frac{1 - \frac{1}{x + 1}}{x})

(0, 0), (5, 4)

What is (\partial)/(\partial y)(x^3+x^2y-y^2-4y) ?
The answer to (\partial)/(\partial y)(x^3+x^2y-y^2-4y) is x^2-2y-4