Precalculus by Richard Wright

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# 12-03 Derivatives

Summary: In this section, you will:

• Find the derivative of a function.
• Find the slope of the tangent line to a function.

SDA NAD Content Standards (2018): PC.4.2, PC.6.4

Velocity is the rate-of-change of position. That means it is how quickly position is changing. Acceleration is the rate-of-change of velocity. In other words, acceleration is how quickly velocity is changing.

Calculus mainly focuses on two problems. The first problem is finding the rate-of-change of functions. This is called finding a derivative. Graphically this is represented by finding the slope of the line tangent to a function. Remember that slopes are rates-of-change. The second problem is finding the area under a curve on a graph. This is called a integral. Integrals also happen to be the opposite of derivatives.

This lesson is about derivatives and rates-of-change. To derive the formula for rate-of-change, start with the slope formula for two points whose x-values are separated by h: (x, f(x)) and (x + h, f(x + h)). See figure 2.

$$Slope = m = \frac{y_2 - y_1}{x_2 - x_1}$$

$$Slope = \frac{f(x + h) - f(x)}{x + h - x}$$

$$Slope = \frac{f(x + h) - f(x)}{h}$$

This gives us the average rate-of-change between two points, but if the instantaneous, or exact, slope at a given point is wanted, h must be zero. So, take the limit as h approaches 0. This produces a function that is called a derivative and is denoted by f′(x). To find the rate-of-change, plug in the x-value of the location of the desired slope.

$$Slope = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$

###### Derivative

$$f′(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$

The rate-of-change at a location is found by plugging in the x-value of the location.

#### Example 1: Find a Derivative

Find the derivative of f(x) = x² + 3

###### Solution

Find f(x + h).

$$f(x + h) = (x + h)^2 + 3$$

Substitute f(x + h) and f(x) into the derivative formula.

$$f′(x) = \lim_{h \rightarrow 0} \frac{\color{blue}{f(x + h)} - \color{red}{f(x)}}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} \frac{(\color{blue}{(x + h)^2 + 3}) - (\color{red}{x^2 + 3})}{h}$$

Simplify the numerator.

$$f′(x) = \lim_{h \rightarrow 0} \frac{(x^2 + 2xh + h^2 + 3) - (x^2 + 3)}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} \frac{2xh + h^2}{h}$$

Factor the h out of the numerator and then cancel the h from the numerator and denominator.

$$f′(x) = \lim_{h \rightarrow 0} \frac{h(2x + h)}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} (2x + h)$$

Now evaluate the limit by plugging in 0 for h.

$$f′(x) = 2x + 0$$

$$f′(x) = 2x$$

#### Example 2: Find a Derivative

Find the derivative of f(x) = x² − 2x.

###### Solution

Find f(x + h).

$$f(x + h) = (x + h)^2 - 2(x + h)$$

Substitute f(x + h) and f(x) into the derivative formula.

$$f′(x) = \lim_{h \rightarrow 0} \frac{\color{blue}{f(x + h)} - \color{red}{f(x)}}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} \frac{(\color{blue}{(x + h)^2 - 2(x + h)}) - (\color{red}{x^2 - 2x})}{h}$$

Simplify the numerator.

$$f′(x) = \lim_{h \rightarrow 0} \frac{(x^2 + 2xh + h^2 - 2x - 2h) - (x^2 - 2x)}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} \frac{2xh + h^2 - 2h}{h}$$

Factor the h out of the numerator and then cancel the h from the numerator and denominator.

$$f′(x) = \lim_{h \rightarrow 0} \frac{h(2x + h - 2)}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} (2x + h - 2)$$

Now evaluate the limit by plugging in 0 for h.

$$f′(x) = 2x + 0 - 2$$

$$f′(x) = 2x - 2$$

##### Try It 1

Find the derivative of f(x) = 2x³.

f ′(x) = 6x²

#### Example 3: Find a Derivative

Find the derivative of $$f(x) = \sqrt{x} - 2$$.

###### Solution

Find f(x + h).

$$f(x + h) = \sqrt{x + h} - 2$$

Substitute f(x + h) and f(x) into the derivative formula.

$$f′(x) = \lim_{h \rightarrow 0} \frac{\color{blue}{f(x + h)} - \color{red}{f(x)}}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} \frac{(\color{blue}{\sqrt{x + h} - 2}) - (\color{red}{\sqrt{x} - 2})}{h}$$

Simplify the numerator.

$$f′(x) = \lim_{h \rightarrow 0} \frac{\sqrt{x + h} - \sqrt{x}}{h}$$

Use the rationalizing technique.

$$f′(x) = \lim_{h \rightarrow 0} \frac{(\sqrt{x + h} - \sqrt{x})}{h} \frac{(\sqrt{x + h} + \sqrt{x})}{(\sqrt{x + h} + \sqrt{x})}$$

$$f′(x) = \lim_{h \rightarrow 0} \frac{(x + h) - (x)}{h(\sqrt{x + h} + \sqrt{x})}$$

$$f′(x) = \lim_{h \rightarrow 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})}$$

Cancel the h from the numerator and denominator.

$$f′(x) = \lim_{h \rightarrow 0} \frac{1}{\sqrt{x + h} + \sqrt{x}}$$

Now evaluate the limit by plugging in 0 for h.

$$f′(x) = \frac{1}{\sqrt{x + 0} + \sqrt{x}}$$

$$f′(x) = \frac{1}{\sqrt{x} + \sqrt{x}}$$

$$f′(x) = \frac{1}{2\sqrt{x}}$$

##### Try It 2

Find the derivative of $$f(x) = \sqrt{x + 1}$$.

$$f′(x) = \frac{1}{2\sqrt{x + 1}}$$

#### Example 4: Find the Slope of a Tangent Line

Find the slope of $$f(x) = x^3$$ at (1, 1).

###### Solution

Start by finding the derivative. So, find f(x + h).

$$f(x + h) = (x + h)^3$$

Substitute f(x + h) and f(x) into the derivative formula.

$$f′(x) = \lim_{h \rightarrow 0} \frac{\color{blue}{f(x + h)} - \color{red}{f(x)}}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} \frac{(\color{blue}{(x + h)^3}) - (\color{red}{x^3})}{h}$$

Simplify the numerator.

$$f′(x) = \lim_{h \rightarrow 0} \frac{(x^3 + 3x^2h + 3xh^2 + h^3) - (x^3)}{h}$$

Factor the h out of the numerator and then cancel the h from the numerator and denominator.

$$f′(x) = \lim_{h \rightarrow 0} \frac{3x^2h + 3xh^2 + h^3}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} \frac{h(3x^2 + 3xh + h^2)}{h}$$

$$f′(x) = \lim_{h \rightarrow 0} 3x^2 + 3xh + h^2$$

Now evaluate the limit by plugging in 0 for h.

$$f′(x) = 3x^2 + 3x(0) + (0)^2$$

$$f′(x) = 3x^2$$

Find the slope at (1, 1) by plugging in x = 1.

$$f′(2) = 3(1)^2 = 3$$

The slope of the tangent line at (1, 1) is 3.

##### Try It 3

Find the slope of $$f(x) = 2x^2 - 4$$ at (–1, –2).

–4

##### Lesson Summary

###### Derivative

$$f′(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}$$

The rate-of-change at a location is found by plugging in the x-value of the location.

## Practice Exercises

1. A function that gives the slope of another function is called a _?_.
2. Find the derivative of the function.

3. $$f(x) = 3x^2$$
4. $$f(x) = (x - 2)^2$$
5. $$f(x) = x^3 + 2x$$
6. $$f(x) = -4x^2 - 2x + 3$$
7. $$f(x) = \sqrt{x - 3}$$
8. $$f(x) = \sqrt{x + 2} - 4$$
9. $$f(x) = \frac{1}{x}$$
10. Find the slope of the function at the given point.

11. $$f(x) = 3x^2 - 4$$ at (2, 8)
12. $$f(x) = 2x^3 + x$$ at (–1, –3)
13. $$f(x) = \sqrt{x}$$ at (4, 2)
14. $$f(x) = \frac{2}{x^2}$$ at (1, 2)
15. $$f(x) = x^2 - 3x + 2$$ at (0, 2)
16. Problem Solving

17. Velocity is the derivative of position with respect to time. A falling object's position can be modeled by $$x(t) = -4.9t^2 + 100$$ where t is time in seconds and x(t) is position in meters. Find the velocity at t = 2 seconds.
18. Acceleration is the derivative of velocity with respect to time. A falling object's velocity can be modeled by $$v(t) = -9.8t$$ where t is time in seconds and v(t) is velocity in meters per second. Find the acceleration at t = 2 seconds.
19. Mixed Review

20. (12-02) Evaluate $$\displaystyle \lim_{x \rightarrow 3} \frac{x^2 - 4x + 3}{x - 3}$$.
21. (12-02) Evaluate $$\displaystyle \lim_{x \rightarrow 0^+} -\frac{4| x |}{x}$$.
22. (12-01) Evaluate $$\displaystyle \lim_{x \rightarrow 2} \cos{πx}$$.
23. (10-02) Use formulas to evaluate $$\displaystyle \sum_{i = 1}^{10} (x^2 - 3x)$$.
24. (10-02) Use formulas to evaluate $$\displaystyle \sum_{i = 1}^{100} (5x^2 - x^3)$$.

1. derivative
2. $$f′(x) = 6x$$
3. $$f′(x) = 2x - 4$$
4. $$f′(x) = 3x^2 + 2$$
5. $$f′(x) = -8x - 2$$
6. $$f′(x) = \frac{1}{2\sqrt{x-3}}$$
7. $$f′(x) = \frac{1}{2\sqrt{x+2}}$$
8. $$f′(x) = -\frac{1}{x^2}$$
9. 12
10. 7
11. $$\frac{1}{4}$$
12. –4
13. –3
14. v = –19.6 m/s
15. a = –9.8 m/s²
16. 2
17. –4
18. 1
19. 220
20. –23810750