Calculus Examples

f (x) = 5 x - 4 x^{2}

$f (x) = 5 x - 4 x^{2}$ at

(- 1, - 9)

$(- 1, - 9)$

Step 1

Find the first derivative and evaluate at

x = - 1

$x = - 1$ and

y = - 9

$y = - 9$ to find the slope of the tangent line.

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Step 1.1

By the Sum Rule, the derivative of

5 x - 4 x^{2}

$5 x - 4 x^{2}$ with respect to

x

$x$ is

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

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

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

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

Step 1.2

Evaluate

\frac{d}{d x} [5 x]

$\frac{d}{d x} [5 x]$ .

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Step 1.2.1

Since

5

$5$ is constant with respect to

x

$x$ , the derivative of

5 x

$5 x$ with respect to

x

$x$ is

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

$5 \frac{d}{d x} [x]$ .

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

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

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

$n = 1$ .

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

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

Step 1.2.3

Multiply

5

$5$ by

1

$1$ .

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

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

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

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

Step 1.3

Evaluate

\frac{d}{d x} [- 4 x^{2}]

$\frac{d}{d x} [- 4 x^{2}]$ .

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Step 1.3.1

Since

- 4

$- 4$ is constant with respect to

x

$x$ , the derivative of

- 4 x^{2}

$- 4 x^{2}$ with respect to

x

$x$ is

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

$- 4 \frac{d}{d x} [x^{2}]$ .

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

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

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 = 2

$n = 2$ .

5 - 4 (2 x)

$5 - 4 (2 x)$

Step 1.3.3

Multiply

2

$2$ by

- 4

$- 4$ .

5 - 8 x

$5 - 8 x$

Step 1.4

Reorder terms.

$- 8 x + 5$

Step 1.5

Evaluate the derivative at

$x = - 1$ .

$- 8 \cdot - 1 + 5$

Step 1.6

Simplify.

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Step 1.6.1

Multiply

$- 8$ by

$- 1$ .

$8 + 5$

Step 1.6.2

Add

$8$ and

$5$ .

$13$

Step 2

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

$y$ .

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Step 2.1

Use the slope

$13$ and a given point

$(- 1, - 9)$ to substitute for

$x_{1}$ and

$y_{1}$ in the point-slope form

$y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 9) = 13 \cdot (x - (- 1))$

Step 2.2

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

$y + 9 = 13 \cdot (x + 1)$

Step 2.3

Solve for

$y$ .

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Step 2.3.1

Simplify

$13 \cdot (x + 1)$ .

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Step 2.3.1.1

Rewrite.

$y + 9 = 0 + 0 + 13 \cdot (x + 1)$

Step 2.3.1.2

Simplify by adding zeros.

$y + 9 = 13 \cdot (x + 1)$

Step 2.3.1.3

Apply the distributive property.

$y + 9 = 13 x + 13 \cdot 1$

Step 2.3.1.4

Multiply

$13$ by

$1$ .

$y + 9 = 13 x + 13$

Step 2.3.2

Move all terms not containing

$y$ to the right side of the equation.

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Step 2.3.2.1

Subtract

$9$ from both sides of the equation.

$y = 13 x + 13 - 9$

Step 2.3.2.2

Subtract

$9$ from

$13$ .

$y = 13 x + 4$

Step 3