Calculus Examples

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

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{3} + 1$ and $g (x) = 3 x^{2} - 4 x + 2$ .

$(x^{3} + 1) \frac{d}{d x} [3 x^{2} - 4 x + 2] + (3 x^{2} - 4 x + 2) \frac{d}{d x} [x^{3} + 1]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $3 x^{2} - 4 x + 2$ with respect to $x$ is $\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [2]$ .

$(x^{3} + 1) (\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [2]) + (3 x^{2} - 4 x + 2) \frac{d}{d x} [x^{3} + 1]$

Step 2.2

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{2}$ with respect to $x$ is $3 \frac{d}{d x} [x^{2}]$ .

$(x^{3} + 1) (3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [2]) + (3 x^{2} - 4 x + 2) \frac{d}{d x} [x^{3} + 1]$

Step 2.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$(x^{3} + 1) (3 (2 x) + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [2]) + (3 x^{2} - 4 x + 2) \frac{d}{d x} [x^{3} + 1]$

Step 2.4

Multiply $2$ by $3$ .

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

Step 2.5

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

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

Step 2.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

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

Step 2.7

Multiply $- 4$ by $1$ .

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

Step 2.8

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$(x^{3} + 1) (6 x - 4 + 0) + (3 x^{2} - 4 x + 2) \frac{d}{d x} [x^{3} + 1]$

Step 2.9

Add $6 x - 4$ and $0$ .

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

Step 2.10

By the Sum Rule, the derivative of $x^{3} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [1]$ .

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

Step 2.11

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

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

Step 2.12

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$(x^{3} + 1) (6 x - 4) + (3 x^{2} - 4 x + 2) (3 x^{2} + 0)$

Step 2.13

Simplify the expression.

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Step 2.13.1

Add $3 x^{2}$ and $0$ .

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

Step 2.13.2

Move $3$ to the left of $3 x^{2} - 4 x + 2$ .

$(x^{3} + 1) (6 x - 4) + 3 \cdot (3 x^{2} - 4 x + 2) x^{2}$

Step 3

Simplify.

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Step 3.1