Calculus Examples

y = 3 x^{3} + 4 x + 5

$y = 3 x^{3} + 4 x + 5$ ,

(2, 37)

$(2, 37)$

Step 1

Find the first derivative and evaluate at

x = 2

$x = 2$ and

y = 37

$y = 37$ to find the slope of the tangent line.

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Step 1.1

By the Sum Rule, the derivative of

3 x^{3} + 4 x + 5

$3 x^{3} + 4 x + 5$ with respect to

x

$x$ is

\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]

$\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]$ .

\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]

$\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]$

Step 1.2

Evaluate

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

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

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Step 1.2.1

Since

3

$3$ is constant with respect to

x

$x$ , the derivative of

3 x^{3}

$3 x^{3}$ with respect to

x

$x$ is

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

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

3 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]

$3 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]$

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

$n = 3$ .

3 (3 x^{2}) + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]

$3 (3 x^{2}) + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]$

Step 1.2.3

Multiply

3

$3$ by

3

$3$ .

9 x^{2} + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]

$9 x^{2} + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]$

9 x^{2} + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]

$9 x^{2} + \frac{d}{d x} [4 x] + \frac{d}{d x} [5]$

Step 1.3

Evaluate

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

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

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Step 1.3.1

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]

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

9 x^{2} + 4 \frac{d}{d x} [x] + \frac{d}{d x} [5]

$9 x^{2} + 4 \frac{d}{d x} [x] + \frac{d}{d x} [5]$

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

$n = 1$ .

9 x^{2} + 4 \cdot 1 + \frac{d}{d x} [5]

$9 x^{2} + 4 \cdot 1 + \frac{d}{d x} [5]$

Step 1.3.3

Multiply

4

$4$ by

1

$1$ .

9 x^{2} + 4 + \frac{d}{d x} [5]

$9 x^{2} + 4 + \frac{d}{d x} [5]$

9 x^{2} + 4 + \frac{d}{d x} [5]

$9 x^{2} + 4 + \frac{d}{d x} [5]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since

5

$5$ is constant with respect to

x

$x$ , the derivative of

5

$5$ with respect to

x

$x$ is

0

$0$ .

9 x^{2} + 4 + 0

$9 x^{2} + 4 + 0$

Step 1.4.2

Add

9 x^{2} + 4

$9 x^{2} + 4$ and

0

$0$ .

9 x^{2} + 4

$9 x^{2} + 4$

9 x^{2} + 4

$9 x^{2} + 4$

Step 1.5

Evaluate the derivative at

x = 2

$x = 2$ .

9 {(2)}^{2} + 4

$9 {(2)}^{2} + 4$

Step 1.6

Simplify.

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Step 1.6.1

Simplify each term.

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Step 1.6.1.1

Raise

2

$2$ to the power of

2

$2$ .

9 \cdot 4 + 4

$9 \cdot 4 + 4$

Step 1.6.1.2

Multiply

9

$9$ by

4

$4$ .

36 + 4

$36 + 4$

36 + 4

$36 + 4$

Step 1.6.2

Add

36

$36$ and

4

$4$ .

40

$40$

40

$40$

40

$40$

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

40

$40$ and a given point

(2, 37)

$(2, 37)$ 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 - (37) = 40 \cdot (x - (2))

$y - (37) = 40 \cdot (x - (2))$

Step 2.2

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

y - 37 = 40 \cdot (x - 2)

$y - 37 = 40 \cdot (x - 2)$

Step 2.3

Solve for

y

$y$ .

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Step 2.3.1

Simplify

40 \cdot (x - 2)

$40 \cdot (x - 2)$ .

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Step 2.3.1.1

Rewrite.

y - 37 = 0 + 0 + 40 \cdot (x - 2)

$y - 37 = 0 + 0 + 40 \cdot (x - 2)$

Step 2.3.1.2

Simplify by adding zeros.

y - 37 = 40 \cdot (x - 2)

$y - 37 = 40 \cdot (x - 2)$

Step 2.3.1.3

Apply the distributive property.

y - 37 = 40 x + 40 \cdot - 2

$y - 37 = 40 x + 40 \cdot - 2$

Step 2.3.1.4

Multiply

40

$40$ by

- 2

$- 2$ .

y - 37 = 40 x - 80

$y - 37 = 40 x - 80$

y - 37 = 40 x - 80

$y - 37 = 40 x - 80$

Step 2.3.2

Move all terms not containing

y

$y$ to the right side of the equation.

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Step 2.3.2.1

Add

37

$37$ to both sides of the equation.

y = 40 x - 80 + 37

$y = 40 x - 80 + 37$

Step 2.3.2.2

Add

- 80

$- 80$ and

37

$37$ .

y = 40 x - 43

$y = 40 x - 43$

y = 40 x - 43

$y = 40 x - 43$

y = 40 x - 43

$y = 40 x - 43$

y = 40 x - 43

$y = 40 x - 43$

Step 3

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