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

The answer to (\partial)/(\partial y)(4x^3y^3-4) is 12x^3y^2

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

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

12 x^{3} y^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (4 x^{3} y^{3}) = 12 x^{3} y^{2}

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

= 12 x^{3} y^{2} - 0

Simplify

= 12 x^{3} y^{2}

y^{^{''}} - 2 y^{^{'}} - 4 y = 0

x \to 2 lim (6 x^{3} + 1)

\frac{d}{d t} (1 - t)

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

\frac{d}{d x} (\frac{( e ^{x} )}{1 + x})

What is (\partial)/(\partial y)(4x^3y^3-4) ?
The answer to (\partial)/(\partial y)(4x^3y^3-4) is 12x^3y^2