# Section 11.4 Question 2

## How do you calculate the derivative of a function from the definition?

The definition of the derivative of a function f (x) at a point x = a was given in Section 11.3,

$$f’\left( a \right) = \mathop {\lim }\limits_{h\,\, \to 0} \frac{{f(a + h) – f(a)}}{h}$$

provided this limit exists. We can adapt this definition to find the derivative of a function by changing the constant a to a variable x.

The derivative of f (x) is defined as $$f’\left( x \right) = \mathop {\lim }\limits_{h\,\, \to 0} \frac{{f(x + h) – f(x)}}{h}$$
provided the limit exists. The symbols f ′(x), are read “f prime of x”.

We can apply this definition in a manner similar to how we applied the definition of derivative of a function at a point.

To find the derivative of f (x) or f ′(x),

1. Evaluate the function f (x) at x + h to give f (x + h). Simplify this expression as much as possible.
2. Form and simplify the difference quotient $\frac{{f(x + h) – f(x)}}{h}$.
3. Take the limit as h approaches 0 of the simplified difference quotient.

Let’s use this strategy to find the derivatives of several functions.

#### Example 2        Find the Derivative

Use the definition of the derivative to find the derivative of the function $$f(x) = 2x – 7$$

Solution To apply the definition of the derivative to this function, we must evaluate f (x + h). Replace x in the function with x + h  to yield $$\displaylines{ f(x + h) = 2\left( {x + h} \right) – 7 \cr = 2x + 2h – 7 \cr}$$

It is important to note that we are replacing x with the group  x + h in parentheses. Now form the difference quotient,$$\frac{{f(x + h) – f(x)}}{h} = \frac{{\left( {2x + 2h – 7} \right) – \left( {2x – 7} \right)}}{h}$$

To make the limit easy to evaluate, let’s simplify the difference quotient as much as possible.

\begin{align*} \frac{{f(x + h) – f(x)}}{h} &= \frac{{\left( {2x + 2h – 7} \right) – \left( {2x – 7} \right)}}{h} && {\color{red}{\small \text{Remove the parentheses and subtract each term}}} \cr &= \frac{{2x + 2h – 7 \color{red}{- 2x + 7}}}{h} && \cr &= \frac{{2 \color{red}{h}}}{\color{red}{h}} && {\color{red}{\small \text{Combine like terms}}} \cr &= 2 && {\color{red}{\small \text{Reduce the quotient}}} \cr \end{align*}

Substitute this expression into the definition of the derivative:

\begin{align*} f'(x) &= \mathop {\lim }\limits_{h\,\, \to 0} \frac{{f(x + h) – f(x)}}{h} \cr &= \mathop {\lim }\limits_{h\,\, \to 0} 2 \cr &= 2 \cr \end{align*}
The derivative of the linear function $f(x) = 2x – 7$ is $f'(x) = 2$.

#### Example 3        Find the Derivative

Use the definition of the derivative to find the derivative of the function$$g(t) = {t^2} + 3t + 7$$

Solution In this example, we’ll follow the same strategy for finding the derivative. However, since the name of the function is g(t), we need to adjust the definition of the derivative to read$$g'(t) = \mathop {\lim }\limits_{h\,\, \to 0} \frac{{g(t + h) – g(t)}}{h}$$

Substitute t + h  for t in g(t) to give
\begin{align*} g(t + h) &= {\left( {t + h} \right)^2} + 3\left( {t + h} \right) + 7 && \cr &= {t^2} + 2ht + {h^{^2}} + 3\left( {t + h} \right) + 7 && \color{red}{\small \text{Multiply }} \color{red}{\left( {t+h} \right)\left( {t+h} \right)} \cr &= {t^2} + 2ht + {h^2} + 3t + 3h + 7 && \color{red}{\small \text{Multiply 3 times }} \color{red}{t+h} \cr \end{align*}

If we place this expression and the expression for  into the numerator of the difference quotient, we get

\begin{align*} \frac{{g(t + h) – g(t)}}{h} &= \frac{{\left( {{t^2} + 2ht + {h^2} + 3t + 3h + 7} \right) – \left( {{t^2} + 3t + 7} \right)}}{h} && \color{red}{\small \text{Subtract and simplify}} \cr &= \frac{{{t^2} + 2ht + {h^2} + 3t + 3h + 7 \color{red}{- {t^2} – 3t – 7}}}{h} && \color{red}{\small \text{Combine like terms}} \cr &= \frac{{2ht + {h^2} + 3h}}{h} && \cr &= \frac{{\color{red}{h}\left( {2t + h + 3} \right)}}{\color{red}{h}} && \color{red}{{\small \text{Factor }}h{\text{ and reduce}}} \cr &= 2t + h + 3 && \cr \end{align*}

With this expression in the definition of the derivative, we write
\begin{align*} g'(t) &= \mathop {\lim }\limits_{h\,\, \to 0} \frac{{g(t + h) – g(t)}}{h} \cr &= \mathop {\lim }\limits_{h\,\, \to 0} \left( {2t + h + 3} \right) \cr &= 2t + 3 \cr \end{align*}

The derivative of the quadratic function $g(t) = {t^2} + 3t + 7$ is the linear function $g'(t) = 2t + 3$.

#### Example 4        Find the Derivative

Use the definition of the derivative to find the derivative of the function$$j(t) = {e^t}$$

Solution The definition of the derivative for this function is$$j'(t) = \mathop {\lim }\limits_{h\,\, \to 0} \frac{{j(t + h) – j(t)}}{h}$$

Using the exponential function, the difference quotient is$$\frac{{j(t + h) – j(t)}}{h} = \frac{{{e^{t + h}} – {e^t}}}{h}$$

This expression can be simplified, but not as easily as Example 2 or Example 3.

In this case we rewrite ${e^{t + h}}$ as ${e^t}\;{e^h}$. By doing this, we can factor the numerator:

\begin{align*} \frac{{j(t + h) – j(t)}}{h} &= \frac{{{e^{t + h}} – {e^t}}}{h} \cr &= \frac{{{e^t}\;{e^h} – {e^t}}}{h} \cr &= \frac{{{e^t}\left( {{e^h} – 1} \right)}}{h} \cr \end{align*}

From this simplified difference quotient, we can write out the definition of the derivative,
\begin{align*} j'(t) &= \mathop {\lim }\limits_{h\,\, \to 0} \frac{{j(t + h) – j(t)}}{h} \cr &= \mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^t}\left( {{e^h} – 1} \right)}}{h} \cr &= {e^t} \cdot \mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^h} – 1}}{h} \cr \end{align*}

The factor ${e^t}$ is a constant with respect to h, so it can be moved outside the limit. To complete the derivative, we must evaluate the limit.

For this limit we’ll find the limit by creating a table of values for  for smaller and smaller h values.

 h $\frac{{{e^h} – 1}}{h}$ 0.1 1.051709181 0.01 1.005016708 0.001 1.000500167 0.0001 1.000050002 0.00001 1.000005000

The table suggests that as x gets smaller and smaller, the value of the quotient approaches 1 or $$\mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^h} – 1}}{h} = 1$$

Using $\mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^h} – 1}}{h} = 1$ in the expression for the definition of the derivative yields

\begin{align*} j'(t) &= {e^t} \cdot \mathop {\lim }\limits_{h\,\, \to 0} \frac{{{e^h} – 1}}{h} \cr &= {e^t} \cdot 1 \cr &= {e^t} \cr \end{align*}

The derivative of $j(t) = {e^t}$  is $j'(t) = {e^t}$.

# How Do You Find The Derivative at a Point From the Definition?

The most difficult part of finding a derivative is evaluating the limit involved in the definition of the derivative at a point. Often there is some algebra and simplifying involved as the example below demonstrates.

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# What is the Difference Between A Secant Line and a Tangent Line?

Many students struggle with slopes of tangent lines versus slopes of secant lines. In the example below, I find these slopes and use them to compute the equation of a tangent line and the equation of a secant line.

The secant line is the red line to the right that passes through two points on the curve. The tangent line is the green line that just grazes the curve at a point.

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# Section 11.4 Question 1

## What is a derivative function?

Let’s examine the tangent line on the function pictured in Figure 1.

Figure 1 – A quadratic function g(x) (blue) with a tangent line (red) at x = 1.

We can approximate the value of the derivative at x = 1 or g′(1) by calculating the slope of the tangent line at x = 1. Any two points on the tangent line can be used to make an estimate of the slope.

Figure 2 – The points (0, 4) and (2, 0) can be used to estimate the slope of the tangent line at x = 1.

From Figure 2, we see that the slope of the tangent line or g′(1) is approximately -4/2 or -2. Any other two points on the line could be used to calculate this slope. For instance, the points (1, 2) and (2, 0) are on the tangent line. The slope between these points is
$$g’\left( 1 \right) \approx \frac{{2 – 0}}{{1 – 2}} = – 2$$

Different points don’t always lead to exactly the same slope, but they should be close. If we look at the tangent line for g(x) at x = 2, we see that it is horizontal and passes through the point (2, 1).

Figure 3 – The tangent line to g(x) at x = 2 is a horizontal tangent line.

Since the tangent line is horizontal, its slope is zero. This tells us that g′(2) = 0.

Let’s estimate the slope of one more tangent line at x = 4.

Figure 4 – The tangent line to g(x) at x = 4 passes through the points (3,1) and (5,9).

The slope is calculated as
$$g’\left( 4 \right) \approx \frac{{9 – 1}}{{5 – 3}} = \frac{8}{2} = 4$$

With these three tangent lines, we can organize the slopes in a table.

If we graph these values with the x values as the independent variable and the slopes as the dependent variable, we see a pattern emerge.

Figure 5 – In this graph, the slopes of the quadratic function are graphed at corresponding x values.

The slopes appear to lie along a straight line. If we draw this line through the points we can use it to find the slope of the tangent line at other x values.

Figure 6 – Each of the slopes of the quadratic function lie along a straight line.

For instance, at x = 3 the line passes though y = 2 . This means the slope of the tangent line at x = 3 on the quadratic function is 2. The function corresponding to this line is called the derivative of the quadratic function g(x) and is denoted by g′(x).

The derivative function g′(x) is a function whose output is the slope of the tangent line to the function y = g(x) at an input x.

All of these concepts can be a bit confusing. So let’s recap these different ideas using a function g(x). Keep in mind that we could use any name in place of g and any variable in place of x.

### Example 1    Find the Derivative Function

The graph of the function f (x) is shown below.

Find the graph of the derivative function f ′(x) using the tangent lines to the function f (x).

Solution To find the derivative of the function, we’ll draw tangent lines along the function, record the slopes in a table and graph the corresponding ordered pairs.

Using the grid on the graph, we see that the tangent line passes through (-3, 0) and (-2, 1.8). The slope of the tangent line is approximately
$$f'( – 3) = \frac{{1.8 – 0}}{{ – 2 – ( – 3)}} = 1.8$$

We’ll continue to fill out the table of tangent line slopes:

Our ability to calculate the slope is constrained by our ability to locate points on the tangent line. For this tangent line, the points (-2, 1) and (1, 1.9) are on the line. The slope of the tangent line is approximately
$$f'( – 2) = \frac{{1.9 – 1}}{{1 – ( – 2)}} = 0.3$$

At the next row in the table, we find the slope of the tangent line at x = -1:

Any two points on the tangent line can be used to find the slope. For this tangent line, the points (-3, 2) and (2, -1) are on the line. The slope of the tangent line is approximately

$$f'( – 1) = \frac{{ – 1 – 2}}{{2 – ( – 3)}} = – 0.6$$

If we continue to find the slopes of tangent lines at each of the x values in the table, we can graph the ordered pairs in the form (x, f ′(x)). This means that we graph the x values horizontally and the f ′(x) values vertically.

These ordered pairs lie on the derivative function f ′(x).

If we were to graph more points on the derivative function, we would see a graph like the one below.

Usually it is not necessary to graph a large number of points to graph the derivative function. Typically a few points on the graph are enough to display the overall pattern. Then the rest of the graph can be sketched to approximate the pattern.

# Section 11.3 Question 4

### What does the derivative at a point tell you about a function?

The derivative at a point is a rate so we may interpret it like any other rate of change. In general, we can use the units on the independent and dependent variable to find the units on the derivative.

Suppose a function is given by y = f (x) where x is the independent variable and y is the dependent variable. The units on the derivative of f (x) at x = a are

Once we have established the units on the derivative, we can use them to interpret what the derivative means.

### Example 5    Find and Interpret the Derivative

Based on data from 2001 through 2010, the annual sales at Apple (in millions of dollars) can be modeled by

where R is the amount spent annually on research and development in millions of dollars.

a.   Find S′(1000).

Solution Using the definition of the derivative at a point, we must compute

We’ll do this by computing S(1000) and S(1000 + h). From these values and expressions we’ll form the difference quotient , simplify, and evaluate the limit.

Let’s look at the terms in the numerator of the difference quotient:

Now form the difference quotient and simplify:

The limit as h approaches 0 yields

The derivative is S′(1000) = 40.780.

b. What does the value you calculated in part a tell you about the relationship between sales and money spent on research and development?

Solution To interpret the derivative in part a, we need to establish the units on the derivative. A derivative has the same units as an average rate of change or an instantaneous rate of change for the function S(R),

In the case of the derivative S′(R), the units are

Notice that the numerator and denominator both are in millions of dollars. To distinguish the dollars of sales from the dollars of research and development, labels indicating the origin of the money are included.

The factor of millions in the numerator and denominator may be divided out to yield

The statement S′(1000) = 40.780 means that at a level of 1000 million dollars of research and development, sales are increasing at a rate of 40.78 dollars of sales per dollar of research and development. In other words, if research and development is increased by 1 dollar at this point, sales will increase by approximately 40.78 dollars.