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

The answer to (\partial)/(\partial x)(4x^2y^3-4) is 8xy^3

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

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

8 x y^{3}

Solution steps

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

\frac{\partial}{\partial x} (4 x^{2} y^{3}) = 8 y^{3} x

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

= 8 y^{3} x - 0

Simplify

= 8 x y^{3}

y^{^{'}} = 0.2 ln (\frac{3000}{y}) y, y (0) = 1000

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

\frac{d}{d x} (π^{2})

x \to 0 lim (8 e^{- x} tan (x))

x \to - 5 lim (\frac{1}{( x ^{2} - 25 )})

What is (\partial)/(\partial x)(4x^2y^3-4) ?
The answer to (\partial)/(\partial x)(4x^2y^3-4) is 8xy^3