2.7: Derivatives and Rates of Change

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Example Video:

Problems

The derivative of $f(x)$ at $x=a$ , denoted $f'(a)$ , is given by $f'(a) = \lim _{x \to a} \frac {f(x)-f(a)}{x-a},$ provided the limit exists. Geometrically, this gives the slope of the tangent line to the graph of $y=f(x)$ at $x=a$ .

An equivalent definition is $\displaystyle f'(a) = \lim _{h \to 0} \frac {f(a+h)-f(a)}{h}$ .

If $f(x) = x^2+x$ :

Compute $f'(2)$ if it exists.
We will use the limit definition. Notice $f(2) = \answer {6}$ . Therefore,
$\begin{eqnarray*} f'(2) &=& \lim_{x \to \answer{2}} \frac{f(x)-f(2)}{x-2} \\ &=& \lim_{x \to 2} \frac{\answer{x^2+x-6}}{x-2}. \end{eqnarray*}$
Notice that if you were to plug in $x=2$ into this limit function, you would get $0/0$ , which means more algebra is needed. In this case, we can , and we can say $x^2+x-6 = (x+3)(\answer {x-2}).$ Therefore, $\lim _{x \to 2} \frac {x^2+x-6}{x-2} = \lim _{x \to 2} \frac {(x+3)(x-2)}{x-2} = \lim _{x \to 2} \answer {x+3} = \answer {5}.$ This tells us $f'(2) = 5$ .
Find the equation of the tangent line to the graph of $y = f(x)$ at $x=2$ .
We have $f'(2)$ which represents of the line. We also know a point on the line, namely $(2,\answer {6})$ . Therefore, we can use point-slope form of the line to get $y-6 = \answer {5}(x-\answer {2}).$ If we wanted to simplify, we could, and we would get $y = \answer {5x-4}.$

If $f(x) = \sqrt {x}$ , then $f'(4) = \answer {1/4}$ .

We use the limit definition again. Notice $f(4) = \answer {2}$ . Therefore, the limit definition becomes

$\begin{eqnarray*} f'(4) &=& \lim_{x \to \answer{4}} \frac{f(x)-f(4)}{x-4} \\ &=& \lim_{x \to 4} \frac{\answer{\sqrt{x}-2}}{x-4}. \end{eqnarray*}$

Notice that if you were to plug in $x=4$ into this limit function, you would get $0/0$ , which means more algebra is needed. In this case, we can factor the numeratormultiply by a conjugate . The conjugate of $\sqrt {x}-2$ is $\answer {\sqrt {x}+2}$ , and $(\sqrt {x}-2)(\sqrt {x}+2) = \answer {x-4}.$ Therefore, the limit becomes $\lim _{x \to 4} \frac {x-4}{(x-4)(\sqrt {x}+2)} = \lim _{x \to 4} \frac {1}{\sqrt {x}+2} = \answer {1/4}.$

A rock is dropped from a $400$ ft cliff. Its height (in feet) at time $t$ (seconds) is given by $h(t) = 400-16t^2.$

When does the rock hit the ground? After $\answer {5}$ feetsecondsfeet per second .
Using the limit definition of the derivative to find the rock’s instantaneous velocity two seconds into its fall, we can say the rock was traveling at $\answer {-64}$ feetsecondsfeet per second .
The answer to the previous part was a negative number. This happened because:
The rock was initially thrown downward. The rock’s height was decreasing two seconds into the trip.
The rock’s average velocity over its entire fall to the ground is equal to $\answer {-80}$ feetsecondsfeet per second .
What is the rock’s instantaneous velocity when it hits the ground? $\answer {-160}$ feetsecondsfeet per second .

Consider $f(x) = x^{1/3}$ and $a = 0$ . We will see that derivatives don’t have to exist.

Calculate $f'(0)$ using the limit definition, if it exists.
First, notice that $f(0) = \answer {0}$ . The limit definition would say $f'(0) = \lim _{x \to 0} \frac {f(x)-f(0)}{x-0} = \lim _{x \to 0} \frac {x^{1/3}}{x} = \lim _{x \to 0} \answer {\frac {1}{x^{2/3}}}.$ This limit does not exist, so $f'(0)$ does not exist.
The limit $\lim _{x \to 0} \frac {1}{x^{2/3}}$ is equal to $\infty$ $-\infty$ neither, because the limit from the left does not match the limit from the right .
What does this tell you about the tangent line to $f(x)$ at $x=0$ ?
It does not exist. The tangent line is vertical. The tangent line is horizontal.