Calculus Examples

f (x) = 3 x^{2} + 2 x

$f (x) = 3 x^{2} + 2 x$ ,

(- 1, 1)

$(- 1, 1)$

Step 1

Find the first derivative and evaluate at

x = - 1

$x = - 1$ and

y = 1

$y = 1$ to find the slope of the tangent line.

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Step 1.1

By the Sum Rule, the derivative of

3 x^{2} + 2 x

$3 x^{2} + 2 x$ with respect to

x

$x$ is

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

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

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

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

Step 1.2

Evaluate

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

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

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Step 1.2.1

Since

3

$3$ is constant with respect to

x

$x$ , the derivative of

3 x^{2}

$3 x^{2}$ with respect to

x

$x$ is

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

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

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

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

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 = 2

$n = 2$ .

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

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

Step 1.2.3

Multiply

2

$2$ by

3

$3$ .

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

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

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

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

Step 1.3

Evaluate

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

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

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Step 1.3.1

Since

2

$2$ is constant with respect to

x

$x$ , the derivative of

2 x

$2 x$ with respect to

x

$x$ is

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

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

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

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

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 \cdot 1

$6 x + 2 \cdot 1$

Step 1.3.3

Multiply

2

$2$ by

1

$1$ .

6 x + 2

$6 x + 2$

6 x + 2

$6 x + 2$

Step 1.4

Evaluate the derivative at

x = - 1

$x = - 1$ .

6 (- 1) + 2

$6 (- 1) + 2$

Step 1.5

Simplify.

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Step 1.5.1

Multiply

6

$6$ by

- 1

$- 1$ .

- 6 + 2

$- 6 + 2$

Step 1.5.2

Add

- 6

$- 6$ and

2

$2$ .

- 4

$- 4$

- 4

$- 4$

- 4

$- 4$

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

- 4

$- 4$ and a given point

(- 1, 1)

$(- 1, 1)$ 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 - (1) = - 4 \cdot (x - (- 1))

$y - (1) = - 4 \cdot (x - (- 1))$

Step 2.2

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

y - 1 = - 4 \cdot (x + 1)

$y - 1 = - 4 \cdot (x + 1)$

Step 2.3

Solve for

y

$y$ .

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Step 2.3.1

Simplify

- 4 \cdot (x + 1)

$- 4 \cdot (x + 1)$ .

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Step 2.3.1.1

Rewrite.

y - 1 = 0 + 0 - 4 \cdot (x + 1)

$y - 1 = 0 + 0 - 4 \cdot (x + 1)$

Step 2.3.1.2

Simplify by adding zeros.

y - 1 = - 4 \cdot (x + 1)

$y - 1 = - 4 \cdot (x + 1)$

Step 2.3.1.3

Apply the distributive property.

y - 1 = - 4 x - 4 \cdot 1

$y - 1 = - 4 x - 4 \cdot 1$

Step 2.3.1.4

Multiply

- 4

$- 4$ by

1

$1$ .

y - 1 = - 4 x - 4

$y - 1 = - 4 x - 4$

y - 1 = - 4 x - 4

$y - 1 = - 4 x - 4$

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

1

$1$ to both sides of the equation.

y = - 4 x - 4 + 1

$y = - 4 x - 4 + 1$

Step 2.3.2.2

Add

- 4

$- 4$ and

1

$1$ .

y = - 4 x - 3

$y = - 4 x - 3$

y = - 4 x - 3

$y = - 4 x - 3$

y = - 4 x - 3

$y = - 4 x - 3$

y = - 4 x - 3

$y = - 4 x - 3$

Step 3