What is the derivative of (x^3-4x^2-6x+1(x^2-3x+1)) ?

The derivative of (x^3-4x^2-6x+1(x^2-3x+1)) is 5x^4-28x^3+21x^2+30x-9

What is the first derivative of (x^3-4x^2-6x+1(x^2-3x+1)) ?

The first derivative of (x^3-4x^2-6x+1(x^2-3x+1)) is 5x^4-28x^3+21x^2+30x-9

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\frac{d}{d x} ((x^{3} - 4 x^{2} - 6 x + 1) (x^{2} - 3 x + 1))

5 x^{4} - 28 x^{3} + 21 x^{2} + 30 x - 9

Solution steps

\frac{d}{d x} ((x^{3} - 4 x^{2} - 6 x + 1) (x^{2} - 3 x + 1))

Apply the Product Rule:

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

f = x^{3} - 4 x^{2} - 6 x + 1, g = x^{2} - 3 x + 1

= \frac{d}{d x} (x^{3} - 4 x^{2} - 6 x + 1) (x^{2} - 3 x + 1) + \frac{d}{d x} (x^{2} - 3 x + 1) (x^{3} - 4 x^{2} - 6 x + 1)

\frac{d}{d x} (x^{3} - 4 x^{2} - 6 x + 1) = 3 x^{2} - 8 x - 6

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

= (3 x^{2} - 8 x - 6) (x^{2} - 3 x + 1) + (2 x - 3) (x^{3} - 4 x^{2} - 6 x + 1)

Simplify

(3 x^{2} - 8 x - 6) (x^{2} - 3 x + 1) + (2 x - 3) (x^{3} - 4 x^{2} - 6 x + 1) : 5 x^{4} - 28 x^{3} + 21 x^{2} + 30 x - 9

= 5 x^{4} - 28 x^{3} + 21 x^{2} + 30 x - 9

What is the derivative of (x^3-4x^2-6x+1(x^2-3x+1)) ?
The derivative of (x^3-4x^2-6x+1(x^2-3x+1)) is 5x^4-28x^3+21x^2+30x-9
What is the first derivative of (x^3-4x^2-6x+1(x^2-3x+1)) ?
The first derivative of (x^3-4x^2-6x+1(x^2-3x+1)) is 5x^4-28x^3+21x^2+30x-9