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

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

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

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

12 x^{2} y^{3}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

x^{3} d y + x y d x = x^{2} d y + 2 y d x

\frac{\partial}{\partial x} (ln (\frac{2 x - 1}{x - 1}))

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

\int sin (t) cos (t) d t

x \to 0.1 lim (\frac{8 sin ( x )}{x})

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