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

The derivative of (x^3-x)(x^2+2)(x^2+x-1) is 7x^6+6x^5+4x^3-9x^2-4x+2

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

The first derivative of (x^3-x)(x^2+2)(x^2+x-1) is 7x^6+6x^5+4x^3-9x^2-4x+2

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derivative of

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

7 x^{6} + 6 x^{5} + 4 x^{3} - 9 x^{2} - 4 x + 2

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

(3 x^{2} - 1) (x^{2} + 2) (x^{2} + x - 1) + (4 x^{3} + 3 x^{2} + 2 x + 2) (x^{3} - x) : 7 x^{6} + 6 x^{5} + 4 x^{3} - 9 x^{2} - 4 x + 2

= 7 x^{6} + 6 x^{5} + 4 x^{3} - 9 x^{2} - 4 x + 2

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = - 2

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

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

\int 4 x \cdot ln (x) d x

f (x) = (\frac{x + 1}{x - 3})^{2}, at x = 7

What is the derivative of (x^3-x)(x^2+2)(x^2+x-1) ?
The derivative of (x^3-x)(x^2+2)(x^2+x-1) is 7x^6+6x^5+4x^3-9x^2-4x+2
What is the first derivative of (x^3-x)(x^2+2)(x^2+x-1) ?
The first derivative of (x^3-x)(x^2+2)(x^2+x-1) is 7x^6+6x^5+4x^3-9x^2-4x+2