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

The derivative of f(x)=(3x^3+2x)(x-2)(x+4) is 15x^4+24x^3-66x^2+8x-16

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

The first derivative of f(x)=(3x^3+2x)(x-2)(x+4) is 15x^4+24x^3-66x^2+8x-16

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derivative of

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

15 x^{4} + 24 x^{3} - 66 x^{2} + 8 x - 16

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d x} (3 x^{3} + 2 x) = 9 x^{2} + 2

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

= (9 x^{2} + 2) (x - 2) (x + 4) + (2 x + 2) (3 x^{3} + 2 x)

Simplify

(9 x^{2} + 2) (x - 2) (x + 4) + (2 x + 2) (3 x^{3} + 2 x) : 15 x^{4} + 24 x^{3} - 66 x^{2} + 8 x - 16

= 15 x^{4} + 24 x^{3} - 66 x^{2} + 8 x - 16

\int \frac{1}{( 4 x ^{2} - 25 ) ^{\frac{3}{2}}} d x

\int - 4 x^{2} - 12 x - 5 d x

x \to 0 + lim (\frac{2}{x})

\int \frac{1}{x ^{π}} d x

x \to 0 lim (\frac{6 x}{3 x})

What is the derivative of f(x)=(3x^3+2x)(x-2)(x+4) ?
The derivative of f(x)=(3x^3+2x)(x-2)(x+4) is 15x^4+24x^3-66x^2+8x-16
What is the first derivative of f(x)=(3x^3+2x)(x-2)(x+4) ?
The first derivative of f(x)=(3x^3+2x)(x-2)(x+4) is 15x^4+24x^3-66x^2+8x-16