Calculus Examples

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

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

(4, 12)

$(4, 12)$

Step 1

Find the first derivative and evaluate at

x = 4

$x = 4$ and

y = 12

$y = 12$ to find the slope of the tangent line.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of

x^{2} - x

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

x

$x$ is

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

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

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

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

Step 1.1.2

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} [- x]

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

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

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

Step 1.2

Evaluate

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

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

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Step 1.2.1

Since

- 1

$- 1$ is constant with respect to

x

$x$ , the derivative of

- x

$- x$ with respect to

x

$x$ is

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

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

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

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

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 = 1

$n = 1$ .

2 x - 1 \cdot 1

$2 x - 1 \cdot 1$

Step 1.2.3

Multiply

- 1

$- 1$ by

1

$1$ .

2 x - 1

$2 x - 1$

2 x - 1

$2 x - 1$

Step 1.3

Evaluate the derivative at

x = 4

$x = 4$ .

2 (4) - 1

$2 (4) - 1$

Step 1.4

Simplify.

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Step 1.4.1

Multiply

2

$2$ by

4

$4$ .

8 - 1

$8 - 1$

Step 1.4.2

Subtract

1

$1$ from

8

$8$ .

7

$7$

7

$7$

7

$7$

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

7

$7$ and a given point

(4, 12)

$(4, 12)$ 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 - (12) = 7 \cdot (x - (4))

$y - (12) = 7 \cdot (x - (4))$

Step 2.2

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

y - 12 = 7 \cdot (x - 4)

$y - 12 = 7 \cdot (x - 4)$

Step 2.3

Solve for

y

$y$ .

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Step 2.3.1

Simplify

7 \cdot (x - 4)

$7 \cdot (x - 4)$ .

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Step 2.3.1.1

Rewrite.

y - 12 = 0 + 0 + 7 \cdot (x - 4)

$y - 12 = 0 + 0 + 7 \cdot (x - 4)$

Step 2.3.1.2

Simplify by adding zeros.

y - 12 = 7 \cdot (x - 4)

$y - 12 = 7 \cdot (x - 4)$

Step 2.3.1.3

Apply the distributive property.

y - 12 = 7 x + 7 \cdot - 4

$y - 12 = 7 x + 7 \cdot - 4$

Step 2.3.1.4

Multiply

7

$7$ by

- 4

$- 4$ .

y - 12 = 7 x - 28

$y - 12 = 7 x - 28$

y - 12 = 7 x - 28

$y - 12 = 7 x - 28$

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

12

$12$ to both sides of the equation.

y = 7 x - 28 + 12

$y = 7 x - 28 + 12$

Step 2.3.2.2

Add

- 28

$- 28$ and

12

$12$ .

y = 7 x - 16

$y = 7 x - 16$

y = 7 x - 16

$y = 7 x - 16$

y = 7 x - 16

$y = 7 x - 16$

y = 7 x - 16

$y = 7 x - 16$

Step 3