Calculus Examples

y = 2 x^{3} - 8 x

$y = 2 x^{3} - 8 x$ ,

(2, 0)

$(2, 0)$

Step 1

Find the first derivative and evaluate at

x = 2

$x = 2$ and

y = 0

$y = 0$ to find the slope of the tangent line.

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Step 1.1

By the Sum Rule, the derivative of

2 x^{3} - 8 x

$2 x^{3} - 8 x$ with respect to

x

$x$ is

\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 8 x]

$\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 8 x]$ .

\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 8 x]

$\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [- 8 x]$

Step 1.2

Evaluate

\frac{d}{d x} [2 x^{3}]

$\frac{d}{d x} [2 x^{3}]$ .

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Step 1.2.1

Since

2

$2$ is constant with respect to

x

$x$ , the derivative of

2 x^{3}

$2 x^{3}$ with respect to

x

$x$ is

2 \frac{d}{d x} [x^{3}]

$2 \frac{d}{d x} [x^{3}]$ .

2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 8 x]

$2 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 8 x]$

Step 1.2.2

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 = 3

$n = 3$ .

2 (3 x^{2}) + \frac{d}{d x} [- 8 x]

$2 (3 x^{2}) + \frac{d}{d x} [- 8 x]$

Step 1.2.3

Multiply

3

$3$ by

2

$2$ .

6 x^{2} + \frac{d}{d x} [- 8 x]

$6 x^{2} + \frac{d}{d x} [- 8 x]$

6 x^{2} + \frac{d}{d x} [- 8 x]

$6 x^{2} + \frac{d}{d x} [- 8 x]$

Step 1.3

Evaluate

\frac{d}{d x} [- 8 x]

$\frac{d}{d x} [- 8 x]$ .

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Step 1.3.1

Since

- 8

$- 8$ is constant with respect to

x

$x$ , the derivative of

- 8 x

$- 8 x$ with respect to

x

$x$ is

- 8 \frac{d}{d x} [x]

$- 8 \frac{d}{d x} [x]$ .

6 x^{2} - 8 \frac{d}{d x} [x]

$6 x^{2} - 8 \frac{d}{d x} [x]$

Step 1.3.2

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

$n = 1$ .

6 x^{2} - 8 \cdot 1

$6 x^{2} - 8 \cdot 1$

Step 1.3.3

Multiply

- 8

$- 8$ by

1

$1$ .

6 x^{2} - 8

$6 x^{2} - 8$

6 x^{2} - 8

$6 x^{2} - 8$

Step 1.4

Evaluate the derivative at

x = 2

$x = 2$ .

6 {(2)}^{2} - 8

$6 {(2)}^{2} - 8$

Step 1.5

Simplify.

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Step 1.5.1

Simplify each term.

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Step 1.5.1.1

Raise

2

$2$ to the power of

2

$2$ .

6 \cdot 4 - 8

$6 \cdot 4 - 8$

Step 1.5.1.2

Multiply

6

$6$ by

4

$4$ .

24 - 8

$24 - 8$

24 - 8

$24 - 8$

Step 1.5.2

Subtract

8

$8$ from

24

$24$ .

16

$16$

16

$16$

16

$16$

Step 2

Plug the slope and point values into the point-slope formula and solve for

y

$y$ .

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Step 2.1

Use the slope

16

$16$ and a given point

(2, 0)

$(2, 0)$ to substitute for

x_{1}

$x_{1}$ and

y_{1}

$y_{1}$ in the point-slope form

y - y_{1} = m (x - x_{1})

$y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

y - (0) = 16 \cdot (x - (2))

$y - (0) = 16 \cdot (x - (2))$

Step 2.2

Simplify the equation and keep it in point-slope form.

y + 0 = 16 \cdot (x - 2)

$y + 0 = 16 \cdot (x - 2)$

Step 2.3

Solve for

y

$y$ .

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Step 2.3.1

Add

y

$y$ and

0

$0$ .

y = 16 \cdot (x - 2)

$y = 16 \cdot (x - 2)$

Step 2.3.2

Simplify

16 \cdot (x - 2)

$16 \cdot (x - 2)$ .

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Step 2.3.2.1

Apply the distributive property.

y = 16 x + 16 \cdot - 2

$y = 16 x + 16 \cdot - 2$

Step 2.3.2.2

Multiply

16

$16$ by

- 2

$- 2$ .

y = 16 x - 32

$y = 16 x - 32$

y = 16 x - 32

$y = 16 x - 32$

y = 16 x - 32

$y = 16 x - 32$

y = 16 x - 32

$y = 16 x - 32$

Step 3