What is the derivative of (5x^3+3)(x^4-2x) ?

The derivative of (5x^3+3)(x^4-2x) is 35x^6-28x^3-6

What is the first derivative of (5x^3+3)(x^4-2x) ?

The first derivative of (5x^3+3)(x^4-2x) is 35x^6-28x^3-6

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derivative of

(5 x^{3} + 3) (x^{4} - 2 x)

35 x^{6} - 28 x^{3} - 6

Solution steps

\frac{d}{d x} ((5 x^{3} + 3) (x^{4} - 2 x))

Apply the Product Rule:

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

f = 5 x^{3} + 3, g = x^{4} - 2 x

= \frac{d}{d x} (5 x^{3} + 3) (x^{4} - 2 x) + \frac{d}{d x} (x^{4} - 2 x) (5 x^{3} + 3)

\frac{d}{d x} (5 x^{3} + 3) = 15 x^{2}

\frac{d}{d x} (x^{4} - 2 x) = 4 x^{3} - 2

= 15 x^{2} (x^{4} - 2 x) + (4 x^{3} - 2) (5 x^{3} + 3)

Simplify

15 x^{2} (x^{4} - 2 x) + (4 x^{3} - 2) (5 x^{3} + 3) : 35 x^{6} - 28 x^{3} - 6

= 35 x^{6} - 28 x^{3} - 6

\int \frac{1}{( 4 x ^{2} + 1 ) ^{2}} d x

x^{2} - 4, [- 4, 3]

\frac{\partial}{\partial x} (x^{2} y + z cos (y))

\frac{d}{d x} (ln (1 + 2 x^{2}))

y^{^{'''}} - 6 y^{^{''}} - y^{^{'}} + 6 y = 0

What is the derivative of (5x^3+3)(x^4-2x) ?
The derivative of (5x^3+3)(x^4-2x) is 35x^6-28x^3-6
What is the first derivative of (5x^3+3)(x^4-2x) ?
The first derivative of (5x^3+3)(x^4-2x) is 35x^6-28x^3-6