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

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

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

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

12 x^{2} y^{2}

Solution steps

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

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

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

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

Simplify

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

\int 12 x^{\frac{5}{7}} + 7 x^{- \frac{6}{7}} d x

y^{^{'}} = \frac{( 3 x ^{2} + y ^{2} )}{x y}

\int co x d x

\int_{1}^{16} \frac{x}{x} d x

y^{^{'}} = (x - 3) e^{- 2 y}

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