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

The answer to (\partial)/(\partial y)(2x-x^3+3xy^2) is 6xy

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

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

6 x y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial y} (3 x y^{2}) = 6 x y

= 0 - 0 + 6 x y

Simplify

= 6 x y

y^{2} (\frac{d x}{d y}) + y^{3} = 1

\int \frac{x ^{2}}{x ^{2} + 1} d x

7 x e^{x}

\int cos (8 x) cos (5 x) d x

\frac{d}{d x} (\frac{5 x}{3})

What is (\partial)/(\partial y)(2x-x^3+3xy^2) ?
The answer to (\partial)/(\partial y)(2x-x^3+3xy^2) is 6xy