Calculus Examples

f (x) = x^{2} - 5

$f (x) = x^{2} - 5$ at

x = 3

$x = 3$

Step 1

Find the corresponding

y

$y$ -value to

x = 3

$x = 3$ .

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Step 1.1

Substitute

3

$3$ in for

x

$x$ .

y = {(3)}^{2} - 5

$y = {(3)}^{2} - 5$

Step 1.2

Solve for

y

$y$ .

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Step 1.2.1

Remove parentheses.

y = 3^{2} - 5

$y = 3^{2} - 5$

Step 1.2.2

Remove parentheses.

y = {(3)}^{2} - 5

$y = {(3)}^{2} - 5$

Step 1.2.3

Simplify

{(3)}^{2} - 5

${(3)}^{2} - 5$ .

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Step 1.2.3.1

Raise

3

$3$ to the power of

2

$2$ .

y = 9 - 5

$y = 9 - 5$

Step 1.2.3.2

Subtract

5

$5$ from

9

$9$ .

y = 4

$y = 4$

y = 4

$y = 4$

y = 4

$y = 4$

y = 4

$y = 4$

Step 2

Find the first derivative and evaluate at

x = 3

$x = 3$ and

y = 4

$y = 4$ to find the slope of the tangent line.

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Step 2.1

By the Sum Rule, the derivative of

x^{2} - 5

$x^{2} - 5$ with respect to

x

$x$ is

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

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 5]$ .

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

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

Step 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 = 2

$n = 2$ .

2 x + \frac{d}{d x} [- 5]

$2 x + \frac{d}{d x} [- 5]$

Step 2.3

Since

- 5

$- 5$ is constant with respect to

x

$x$ , the derivative of

- 5

$- 5$ with respect to

x

$x$ is

0

$0$ .

2 x + 0

$2 x + 0$

Step 2.4

Add

2 x

$2 x$ and

0

$0$ .

2 x

$2 x$

Step 2.5

Evaluate the derivative at

x = 3

$x = 3$ .

2 (3)

$2 (3)$

Step 2.6

Multiply

2

$2$ by

3

$3$ .

6

$6$

6

$6$

Step 3

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

y

$y$ .

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Step 3.1

Use the slope

6

$6$ and a given point

(3, 4)

$(3, 4)$ 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 - (4) = 6 \cdot (x - (3))

$y - (4) = 6 \cdot (x - (3))$

Step 3.2

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

y - 4 = 6 \cdot (x - 3)

$y - 4 = 6 \cdot (x - 3)$

Step 3.3

Solve for

y

$y$ .

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Step 3.3.1

Simplify

6 \cdot (x - 3)

$6 \cdot (x - 3)$ .

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Step 3.3.1.1

Rewrite.

y - 4 = 0 + 0 + 6 \cdot (x - 3)

$y - 4 = 0 + 0 + 6 \cdot (x - 3)$

Step 3.3.1.2

Simplify by adding zeros.

y - 4 = 6 \cdot (x - 3)

$y - 4 = 6 \cdot (x - 3)$

Step 3.3.1.3

Apply the distributive property.

y - 4 = 6 x + 6 \cdot - 3

$y - 4 = 6 x + 6 \cdot - 3$

Step 3.3.1.4

Multiply

6

$6$ by

- 3

$- 3$ .

y - 4 = 6 x - 18

$y - 4 = 6 x - 18$

y - 4 = 6 x - 18

$y - 4 = 6 x - 18$

Step 3.3.2

Move all terms not containing

y

$y$ to the right side of the equation.

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Step 3.3.2.1

Add

4

$4$ to both sides of the equation.

y = 6 x - 18 + 4

$y = 6 x - 18 + 4$

Step 3.3.2.2

Add

- 18

$- 18$ and

4

$4$ .

y = 6 x - 14

$y = 6 x - 14$

y = 6 x - 14

$y = 6 x - 14$

y = 6 x - 14

$y = 6 x - 14$

y = 6 x - 14

$y = 6 x - 14$

Step 4