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

The answer to (\partial)/(\partial x)((2x^3-4y)(2x^2-4)) is 20x^4-24x^2-16xy

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

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

20 x^{4} - 24 x^{2} - 16 x y

Solution steps

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

Treat

y

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 2 x^{3} - 4 y, g = 2 x^{2} - 4

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

\frac{\partial}{\partial x} (2 x^{3} - 4 y) = 6 x^{2}

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

= 6 x^{2} (2 x^{2} - 4) + 4 x (2 x^{3} - 4 y)

Simplify

6 x^{2} (2 x^{2} - 4) + 4 x (2 x^{3} - 4 y) : 20 x^{4} - 24 x^{2} - 16 x y

= 20 x^{4} - 24 x^{2} - 16 x y

\frac{d}{d x} (cos (\frac{π}{2}))

\int \frac{1}{x ^{2} - 2 x + 8} d x

\int \frac{1}{3 x} d x

\int (tan (x))^{3} (sec (x))^{2} d x

\frac{1}{x ^{4}}

What is (\partial)/(\partial x)((2x^3-4y)(2x^2-4)) ?
The answer to (\partial)/(\partial x)((2x^3-4y)(2x^2-4)) is 20x^4-24x^2-16xy