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

The answer to (\partial)/(\partial y)(2xy^3-4) is 6xy^2

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(\partial)/(\partial y)(2xy^3-4)

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

6 x y^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 6 x y^{2} - 0

Simplify

= 6 x y^{2}

y^{^{''}} - 18 y^{^{'}} + 81 y = 0

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What is (\partial)/(\partial y)(2xy^3-4) ?
The answer to (\partial)/(\partial y)(2xy^3-4) is 6xy^2