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

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

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

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

6 x y^{2}

Solution steps

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

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} (3)

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

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

= 6 x y^{2} - 0

Simplify

= 6 x y^{2}

4 x^{\frac{5}{4}} + 8 x^{\frac{3}{2}} + 3 x

y^{^{'}} = \frac{( y - y ^{2} )}{x}

\frac{1 + x}{1 - x} - 1

f (x) = sec (x^{2})

\frac{d}{d x} (9 + 3 x + sin (3 x))

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