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

The derivative of (x^3-2x+1(2x^2+3x)) is 10x^4+12x^3-12x^2-8x+3

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

The first derivative of (x^3-2x+1(2x^2+3x)) is 10x^4+12x^3-12x^2-8x+3

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

10 x^{4} + 12 x^{3} - 12 x^{2} - 8 x + 3

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

(3 x^{2} - 2) (2 x^{2} + 3 x) + (4 x + 3) (x^{3} - 2 x + 1) : 10 x^{4} + 12 x^{3} - 12 x^{2} - 8 x + 3

= 10 x^{4} + 12 x^{3} - 12 x^{2} - 8 x + 3

f (x) = \frac{x - 6}{9 x - 1}, at x = 1

m y d x = n x d y

\int_{0}^{1} \frac{1}{4 + x ^{2}} d x

x \to 0 lim ((\frac{1}{x} + 1)^{x})

What is the derivative of (x^3-2x+1(2x^2+3x)) ?
The derivative of (x^3-2x+1(2x^2+3x)) is 10x^4+12x^3-12x^2-8x+3
What is the first derivative of (x^3-2x+1(2x^2+3x)) ?
The first derivative of (x^3-2x+1(2x^2+3x)) is 10x^4+12x^3-12x^2-8x+3