Calculus Examples

y = 3 x^{2} - 2 x + 1

$y = 3 x^{2} - 2 x + 1$ ,

(1, 2)

$(1, 2)$

Step 1

Find the first derivative and evaluate at

x = 1

$x = 1$ and

y = 2

$y = 2$ 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 + 1

$3 x^{2} - 2 x + 1$ with respect to

x

$x$ is

\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]

$\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$ .

\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]

$\frac{d}{d x} [3 x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$

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] + \frac{d}{d x} [1]

$3 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$

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] + \frac{d}{d x} [1]

$3 (2 x) + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$

Step 1.2.3

Multiply

2

$2$ by

3

$3$ .

6 x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]

$6 x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$

6 x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]

$6 x + \frac{d}{d x} [- 2 x] + \frac{d}{d x} [1]$

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] + \frac{d}{d x} [1]

$6 x - 2 \frac{d}{d x} [x] + \frac{d}{d x} [1]$

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 + \frac{d}{d x} [1]

$6 x - 2 \cdot 1 + \frac{d}{d x} [1]$

Step 1.3.3

Multiply

- 2

$- 2$ by

1

$1$ .

6 x - 2 + \frac{d}{d x} [1]

$6 x - 2 + \frac{d}{d x} [1]$

6 x - 2 + \frac{d}{d x} [1]

$6 x - 2 + \frac{d}{d x} [1]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since

1

$1$ is constant with respect to

x

$x$ , the derivative of

1

$1$ with respect to

x

$x$ is

0

$0$ .

6 x - 2 + 0

$6 x - 2 + 0$

Step 1.4.2

Add

6 x - 2

$6 x - 2$ and

0

$0$ .

6 x - 2

$6 x - 2$

6 x - 2

$6 x - 2$

Step 1.5

Evaluate the derivative at

x = 1

$x = 1$ .

6 (1) - 2

$6 (1) - 2$

Step 1.6

Simplify.

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Step 1.6.1

Multiply

6

$6$ by

1

$1$ .

6 - 2

$6 - 2$

Step 1.6.2

Subtract

2

$2$ from

6

$6$ .

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, 2)

$(1, 2)$ 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 - (2) = 4 \cdot (x - (1))

$y - (2) = 4 \cdot (x - (1))$

Step 2.2

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

y - 2 = 4 \cdot (x - 1)

$y - 2 = 4 \cdot (x - 1)$

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 - 2 = 0 + 0 + 4 \cdot (x - 1)

$y - 2 = 0 + 0 + 4 \cdot (x - 1)$

Step 2.3.1.2

Simplify by adding zeros.

y - 2 = 4 \cdot (x - 1)

$y - 2 = 4 \cdot (x - 1)$

Step 2.3.1.3

Apply the distributive property.

y - 2 = 4 x + 4 \cdot - 1

$y - 2 = 4 x + 4 \cdot - 1$

Step 2.3.1.4

Multiply

4

$4$ by

- 1

$- 1$ .

y - 2 = 4 x - 4

$y - 2 = 4 x - 4$

y - 2 = 4 x - 4

$y - 2 = 4 x - 4$

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

2

$2$ to both sides of the equation.

y = 4 x - 4 + 2

$y = 4 x - 4 + 2$

Step 2.3.2.2

Add

- 4

$- 4$ and

2

$2$ .

y = 4 x - 2

$y = 4 x - 2$

y = 4 x - 2

$y = 4 x - 2$

y = 4 x - 2

$y = 4 x - 2$

y = 4 x - 2

$y = 4 x - 2$

Step 3