What is the derivative of (x^2+5x+1(x^3-2x)) ?

The derivative of (x^2+5x+1(x^3-2x)) is 5x^4+20x^3-3x^2-20x-2

What is the first derivative of (x^2+5x+1(x^3-2x)) ?

The first derivative of (x^2+5x+1(x^3-2x)) is 5x^4+20x^3-3x^2-20x-2

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\frac{d}{d x} ((x^{2} + 5 x + 1) (x^{3} - 2 x))

5 x^{4} + 20 x^{3} - 3 x^{2} - 20 x - 2

Solution steps

\frac{d}{d x} ((x^{2} + 5 x + 1) (x^{3} - 2 x))

Apply the Product Rule:

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

f = x^{2} + 5 x + 1, g = x^{3} - 2 x

= \frac{d}{d x} (x^{2} + 5 x + 1) (x^{3} - 2 x) + \frac{d}{d x} (x^{3} - 2 x) (x^{2} + 5 x + 1)

\frac{d}{d x} (x^{2} + 5 x + 1) = 2 x + 5

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

= (2 x + 5) (x^{3} - 2 x) + (3 x^{2} - 2) (x^{2} + 5 x + 1)

Simplify

(2 x + 5) (x^{3} - 2 x) + (3 x^{2} - 2) (x^{2} + 5 x + 1) : 5 x^{4} + 20 x^{3} - 3 x^{2} - 20 x - 2

= 5 x^{4} + 20 x^{3} - 3 x^{2} - 20 x - 2

y^{^{'}} + 4 y = - 1

\frac{d y}{d x} = \frac{y - y}{tan ( x )}

\int 8 sec^{3} (5 x) d x

x \to 7 + lim (\frac{5}{x - 7})

y^{^{'}} (x^{2} + 1) = y

What is the derivative of (x^2+5x+1(x^3-2x)) ?
The derivative of (x^2+5x+1(x^3-2x)) is 5x^4+20x^3-3x^2-20x-2
What is the first derivative of (x^2+5x+1(x^3-2x)) ?
The first derivative of (x^2+5x+1(x^3-2x)) is 5x^4+20x^3-3x^2-20x-2