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3.1 Introducing the Derivative

Graph of a function y=f(x) showing a secant line through (a, f(a)) and (x, f(x)), and a tangent line at (a, f(a)).

The graph illustrates a function curve \( y = f(x) \) with two specific lines: a secant line passing through points \( (a, f(a)) \) and \( (x, f(x)) \), and a tangent line at the point \( (a, f(a)) \).

Slopes and Definitions

slope of secant line:

\[ \frac{f(x) - f(a)}{x - a} \]

slope of tangent line at \( x = a \)

  • tangent line: touches only \( (a, f(a)) \)
  • slope of tangent line can't be found using basic geometry but we can approximate it by using a secant line whose two points are close together
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Graph showing multiple secant lines approaching the tangent line as the point x moves closer to a.

As the point \( x \) is brought closer to \( a \), the secant line becomes more like the tangent line. This process is represented by the limit as \( x \to a \).

The Formal Definition

tangent line slope of \( f(x) \) at \( x = a \) is:

\[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \]

"f prime of a"

this is also called the derivative of \( f(x) \) at \( x = a \)

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Another form of the derivative formula

Consider a function curve \( y = f(x) \) on a coordinate plane. We can define the slope of a secant line passing through two points on this curve.

Graph showing a curve y=f(x) with points at a and a+h, showing secant and tangent lines.
\[ \text{secant line slope} = \frac{f(a+h) - f(a)}{(a+h) - a} = \frac{f(a+h) - f(a)}{h} \]

The distance between the two points on the x-axis is \( h \).

To find the slope of the tangent line, we make the second point come to the first: shrink \( h \to 0 \), which means taking the limit as \( h \to 0 \).

So, the tangent line slope at \( x = a \) is:

\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]
  • Secant line slope: average rate of change over an interval
  • Tangent line slope: instantaneous rate of change at \( x = a \) or over a very short interval (\( x \to a \) or \( h \to 0 \))
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Example

Given \( f(x) = \frac{1}{4+3x} \), find \( f'(2) \). Here, \( a = 2 \).

We can use either form of \( f'(a) \). Let's use the 2nd form in this example:

\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \quad \text{here, } a = 2 \]
\[ f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h} \]

Note: if \( h = 0 \), then \( \frac{f(2) - f(2)}{0} \to \frac{0}{0} = ? \)

\[ = \lim_{h \to 0} \frac{\frac{1}{4+3(2+h)} - \frac{1}{4+3(2)}}{h} \]

Combine the top:

\[ = \lim_{h \to 0} \frac{\frac{1}{10+3h} - \frac{1}{10}}{h} = \lim_{h \to 0} \frac{\frac{1}{10+3h} \cdot \frac{10}{10} - \frac{1}{10} \cdot \frac{10+3h}{10+3h}}{h} \]
\[ = \lim_{h \to 0} \frac{\frac{10}{10(10+3h)} - \frac{10+3h}{10(10+3h)}}{h} \]
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\[ = \lim_{h \to 0} \frac{\frac{10 - (10 + 3h)}{10(10 + 3h)}}{h} = \lim_{h \to 0} \left( \frac{-3h}{10(10 + 3h)} \cdot \frac{1}{h} \right) \]
\[ = \lim_{h \to 0} \frac{-3}{10(10 + 3h)} = \boxed{-\frac{3}{100}} = f'(2) \]

Slope of tangent line of \( f(x) = \frac{1}{4 + 3x} \) at \( x = 2 \)

Equation of tangent line?

It is a line through \( (2, f(2)) = (2, \frac{1}{10}) \) with slope \( f'(2) = -\frac{3}{100} \)

Point-slope form: \( y - y_1 = m(x - x_1) \)

\[ y - \frac{1}{10} = -\frac{3}{100}(x - 2) \]
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Example

\( f(x) = \sqrt{x - 2} \) find \( f'(3) \)

This time, use \( f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \)

Here, \( a = 3 \)

\[ f'(3) = \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} \]
\[ = \lim_{x \to 3} \frac{\sqrt{x - 2} - 1}{x - 3} \to \frac{0}{0} \text{ as } x = 3 \]

We see root so rationalize

\[ = \lim_{x \to 3} \frac{\sqrt{x - 2} - 1}{x - 3} \cdot \frac{\sqrt{x - 2} + 1}{\sqrt{x - 2} + 1} \]
\[ = \lim_{x \to 3} \frac{(\sqrt{x - 2})^2 - 1^2}{(x - 3)(\sqrt{x - 2} + 1)} = \lim_{x \to 3} \frac{(x - 3)}{(x - 3)(\sqrt{x - 2} + 1)} \]

Now let \( x = 3 \)

\[ f'(3) = \frac{1}{2} \]
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Example: Identifying Functions from the Derivative Definition

If

\[ f'(a) = \lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h} \]

what is \( f(x) \) and what is \( a \)?

First, recognize the form of the derivative. We see \( \lim_{h \to 0} \) so we must be using:

\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]

Comparing the two expressions:

\[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \]\[ f'(a) = \lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h} \]

Note that \( 3 \) can be written as \( \sqrt{9} \).

Line them up:

  • \( f(a+h) = \sqrt{9+h} \)
  • \( f(a) = \sqrt{9} = 3 \)

Now think of a \( f(x) \) and a pair to make them work.

Try: \( f(x) = \sqrt{x} \) and \( a = 9 \) \( \rightarrow f(a) = \sqrt{a} = f(9) = \sqrt{9} = 3 \)

\( f(a+h) = f(9+h) = \sqrt{9+h} \)

So, \( f(x) = \sqrt{x} \) and \( a = 9 \)
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Population Growth and Rates of Change

Graph of Population (thousands) vs Years after 1950, showing an exponential curve with tangent and secant lines.

The graph illustrates population growth over time. A tangent line is shown at \( t = 30 \), and a secant line is shown between \( t = 20 \) and \( t = 30 \).

Year195019601970198019902000
t01020304050
p(t)59,000114,153220,862427,322826,7791,599,646

Sometimes we can estimate the instantaneous rate of change by looking at the average rate of change.

  • avg rate: secant line slope
  • inst rate: tangent line slope
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Estimating Instantaneous Rate of Change

Notice the slopes of the secant line on \( 20 \le t \le 30 \) and the tangent line slope at \( t = 30 \) are close.

Since we don't know what \( P(t) \) looks like, we can't find \( P'(30) \) but by comparing the slopes we can say:

\[ P'(30) \approx \text{avg. rate of change over } 20 \le t \le 30 \]
\[ \approx \frac{P(30) - P(20)}{30 - 20} \approx \frac{427322 - 220862}{10} \approx 20652 \]