Calculus Examples

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

${(2 x + 1)}^{2}$

Step 1

Rewrite

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

${(2 x + 1)}^{2}$ as

(2 x + 1) (2 x + 1)

$(2 x + 1) (2 x + 1)$ .

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

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

Step 2

Expand

(2 x + 1) (2 x + 1)

$(2 x + 1) (2 x + 1)$ using the FOIL Method.

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Step 2.1

Apply the distributive property.

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

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

Step 2.2

Apply the distributive property.

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

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

Step 2.3

Apply the distributive property.

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

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

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

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

Step 3

Simplify and combine like terms.

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Step 3.1

Simplify each term.

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Step 3.1.1

Rewrite using the commutative property of multiplication.

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

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

Step 3.1.2

Multiply

x

$x$ by

x

$x$ by adding the exponents.

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Step 3.1.2.1

Move

x

$x$ .

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

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

Step 3.1.2.2

Multiply

x

$x$ by

x

$x$ .

\frac{d}{d x} [2 \cdot 2 x^{2} + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1]

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

\frac{d}{d x} [2 \cdot 2 x^{2} + 2 x \cdot 1 + 1 (2 x) + 1 \cdot 1]

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

Step 3.1.3

Multiply

2

$2$ by

2

$2$ .

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

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

Step 3.1.4

Multiply

2

$2$ by

1

$1$ .

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

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

Step 3.1.5

Multiply

2 x

$2 x$ by

1

$1$ .

\frac{d}{d x} [4 x^{2} + 2 x + 2 x + 1 \cdot 1]

$\frac{d}{d x} [4 x^{2} + 2 x + 2 x + 1 \cdot 1]$

Step 3.1.6

Multiply

1

$1$ by

1

$1$ .

\frac{d}{d x} [4 x^{2} + 2 x + 2 x + 1]

$\frac{d}{d x} [4 x^{2} + 2 x + 2 x + 1]$

\frac{d}{d x} [4 x^{2} + 2 x + 2 x + 1]

$\frac{d}{d x} [4 x^{2} + 2 x + 2 x + 1]$

Step 3.2

Add

2 x

$2 x$ and

2 x

$2 x$ .

\frac{d}{d x} [4 x^{2} + 4 x + 1]

$\frac{d}{d x} [4 x^{2} + 4 x + 1]$

\frac{d}{d x} [4 x^{2} + 4 x + 1]

$\frac{d}{d x} [4 x^{2} + 4 x + 1]$

Step 4

By the Sum Rule, the derivative of

4 x^{2} + 4 x + 1

$4 x^{2} + 4 x + 1$ with respect to

x

$x$ is

\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [1]

$\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [1]$ .

\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [1]

$\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [1]$

Step 5

Since

4

$4$ is constant with respect to

x

$x$ , the derivative of

4 x^{2}

$4 x^{2}$ with respect to

x

$x$ is

4 \frac{d}{d x} [x^{2}]

$4 \frac{d}{d x} [x^{2}]$ .

4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [1]

$4 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [1]$

Step 6

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 2

$n = 2$ .

4 (2 x) + \frac{d}{d x} [4 x] + \frac{d}{d x} [1]

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

Step 7

Multiply

2

$2$ by

4

$4$ .

8 x + \frac{d}{d x} [4 x] + \frac{d}{d x} [1]

$8 x + \frac{d}{d x} [4 x] + \frac{d}{d x} [1]$

Step 8

Since

4

$4$ is constant with respect to

x

$x$ , the derivative of

4 x

$4 x$ with respect to

x

$x$ is

4 \frac{d}{d x} [x]

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

8 x + 4 \frac{d}{d x} [x] + \frac{d}{d x} [1]

$8 x + 4 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

Step 9

Differentiate using the Power Rule which states that

\frac{d}{d x} [x^{n}]

$\frac{d}{d x} [x^{n}]$ is

n x^{n - 1}

$n x^{n - 1}$ where

n = 1

$n = 1$ .

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

$8 x + 4 \cdot 1 + \frac{d}{d x} [1]$

Step 10

Multiply

4

$4$ by

1

$1$ .

8 x + 4 + \frac{d}{d x} [1]

$8 x + 4 + \frac{d}{d x} [1]$

Step 11

Since

1

$1$ is constant with respect to

x

$x$ , the derivative of

1

$1$ with respect to

x

$x$ is

0

$0$ .

8 x + 4 + 0

$8 x + 4 + 0$

Step 12

Add

8 x + 4

$8 x + 4$ and

0

$0$ .

8 x + 4

$8 x + 4$