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

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

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

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

2 x - 3 y^{2}

Solution steps

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

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

\frac{\partial}{\partial y} (y^{3}) = 3 y^{2}

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

= 0 - 3 y^{2} + 2 x

Simplify

= 2 x - 3 y^{2}

cos (t) \cdot cos (π)

\int sin^{5} (3 x) cos (3 x) d x

f (x) = x^{2} - 5, (- 4, 11)

\int_{0}^{10} 1000 t e^{- 0.5 t} d t

\frac{d}{d θ} (e^{i θ})

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