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

The derivative of (x^3+4x)(3x^2+2x-5) is 15x^4+8x^3+21x^2+16x-20

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

The first derivative of (x^3+4x)(3x^2+2x-5) is 15x^4+8x^3+21x^2+16x-20

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derivative of

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

15 x^{4} + 8 x^{3} + 21 x^{2} + 16 x - 20

Solution steps

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

Apply the Product Rule:

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

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

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

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

\frac{d}{d x} (3 x^{2} + 2 x - 5) = 6 x + 2

= (3 x^{2} + 4) (3 x^{2} + 2 x - 5) + (6 x + 2) (x^{3} + 4 x)

Simplify

(3 x^{2} + 4) (3 x^{2} + 2 x - 5) + (6 x + 2) (x^{3} + 4 x) : 15 x^{4} + 8 x^{3} + 21 x^{2} + 16 x - 20

= 15 x^{4} + 8 x^{3} + 21 x^{2} + 16 x - 20

\int m + n y d y

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

x \to e lim (ln ∣ 3 x ∣)

25 y^{^{''}} - y = x e^{\frac{x}{5}}

1 e^{x}

What is the derivative of (x^3+4x)(3x^2+2x-5) ?
The derivative of (x^3+4x)(3x^2+2x-5) is 15x^4+8x^3+21x^2+16x-20
What is the first derivative of (x^3+4x)(3x^2+2x-5) ?
The first derivative of (x^3+4x)(3x^2+2x-5) is 15x^4+8x^3+21x^2+16x-20