Calculus Examples

$f (x) = (x - 3) (2 x + 2)$

Step 1

Find the first derivative.

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Step 1.1

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 $f (x) = x - 3$ and $g (x) = 2 x + 2$ .

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

Step 1.2

Differentiate.

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Step 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.2.4

Multiply $2$ by $1$ .

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

Step 1.2.5

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

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

Step 1.2.6

Simplify the expression.

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Step 1.2.6.1

Add $2$ and $0$ .

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

Step 1.2.6.2

Move $2$ to the left of $x - 3$ .

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

Step 1.2.7

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

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

Step 1.2.8

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

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

Step 1.2.9

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3$ with respect to $x$ is $0$ .

$2 (x - 3) + (2 x + 2) (1 + 0)$

Step 1.2.10

Simplify the expression.

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Step 1.2.10.1

Add $1$ and $0$ .

$2 (x - 3) + (2 x + 2) \cdot 1$

Step 1.2.10.2

Multiply $2 x + 2$ by $1$ .

$2 (x - 3) + 2 x + 2$

Step 1.3

Simplify.

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Step 1.3.1

Apply the distributive property.

$2 x + 2 \cdot - 3 + 2 x + 2$

Step 1.3.2

Combine terms.

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Step 1.3.2.1

Multiply $2$ by $- 3$ .

$2 x - 6 + 2 x + 2$

Step 1.3.2.2

Add $2 x$ and $2 x$ .

$4 x - 6 + 2$

Step 1.3.2.3

Add $- 6$ and $2$ .

$f' (x) = 4 x - 4$

Step 2

Find the second derivative.

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Step 2.1

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

$\frac{d}{d x} [4 x] + \frac{d}{d x} [- 4]$

Step 2.2

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 2.2.1

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

$4 \frac{d}{d x} [x] + \frac{d}{d x} [- 4]$

Step 2.2.2

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

$4 \cdot 1 + \frac{d}{d x} [- 4]$

Step 2.2.3

Multiply $4$ by $1$ .

$4 + \frac{d}{d x} [- 4]$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$4 + 0$

Step 2.3.2

Add $4$ and $0$ .

$f'' (x) = 4$

Step 3

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$f''' (x) = 0$

Step 4

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

$f^{4} (x) = 0$

Step 5

The fourth derivative of $f (x)$ with respect to $x$ is $0$ .

$0$