What is the derivative of (3x^2+9)(x^2-1x) ?

The derivative of (3x^2+9)(x^2-1x) is 12x^3-9x^2+18x-9

What is the first derivative of (3x^2+9)(x^2-1x) ?

The first derivative of (3x^2+9)(x^2-1x) is 12x^3-9x^2+18x-9

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derivative of

(3 x^{2} + 9) (x^{2} - 1 x)

12 x^{3} - 9 x^{2} + 18 x - 9

Solution steps

\frac{d}{d x} ((3 x^{2} + 9) (x^{2} - 1 \cdot x))

Simplify

(3 x^{2} + 9) (x^{2} - 1 \cdot x) : (3 x^{2} + 9) (x^{2} - x)

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

Apply the Product Rule:

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

f = 3 x^{2} + 9, g = x^{2} - x

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

\frac{d}{d x} (3 x^{2} + 9) = 6 x

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

= 6 x (x^{2} - x) + (2 x - 1) (3 x^{2} + 9)

Simplify

6 x (x^{2} - x) + (2 x - 1) (3 x^{2} + 9) : 12 x^{3} - 9 x^{2} + 18 x - 9

= 12 x^{3} - 9 x^{2} + 18 x - 9

(2 x y - 9 x^{2}) + (2 y + x^{2} + 1) \frac{d y}{d x} = 0

s l o p e f (x) = - 8 x^{2} + 8

a re a y = x - 1, y = 0, x = 9

d er i v a t i v e f (x) = x^{e}

\int_{- \infty}^{0} \frac{1}{6 - 9 x} d x

What is the derivative of (3x^2+9)(x^2-1x) ?
The derivative of (3x^2+9)(x^2-1x) is 12x^3-9x^2+18x-9
What is the first derivative of (3x^2+9)(x^2-1x) ?
The first derivative of (3x^2+9)(x^2-1x) is 12x^3-9x^2+18x-9