Precalculus by Richard Wright

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12-03 Derivatives

Mr. Wright teaches the lesson.

Summary: In this section, you will:

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

velocity
(Pixabay/PublicDomainPictures)

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.

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
Finding the slope between two points.

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

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

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

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.

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

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

Find the derivative of f(x) = 2x³.

Answer

f ′(x) = 6x2

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.

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

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

$$ f′(x) = \frac{1}{2\sqrt{x}} $$

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

Answer

\(f′(x) = \frac{1}{2\sqrt{x + 1}}\)

Find the Slope of a Tangent Line

Find the slope of f(x) = x3 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) = 3x2 + 3x(0) + (0)2

f ′(x) = 3x2

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.

f(x) and tangent line
\(f(x) = x^3\) and the tangent line at (1, 1) whose slope is 3.

Find the slope of f(x) = 2x2 − 4 at (–1, –2).

Answer

–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.

Helpful videos about this lesson.

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) = 3x2
  4. f(x) = (x − 2)2
  5. f(x) = x3 + 2x
  6. f(x) = −4x2 − 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) = 3x2 − 4 at (2, 8)
  12. f(x) = 2x3 + 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) = x2 − 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) = −4.9t2 + 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 \(\displaystyle \sum_{i = 1}^{100} (5x^2 - x^3)\).

Answers

  1. derivative
  2. f ′(x) = 6x
  3. f ′(x) = 2x − 4
  4. f ′(x) = 3x2 + 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