2.8: Derivatives as Functions

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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

If $f(x)$ is a function, the derivative, denoted $f'(x)$ or $\displaystyle \frac {dy}{dx}$ , is defined as

$f'(x) = \lim _{h \to 0} \frac {f(x+h)-f(x)}{h},$ provided the limit exists.

If $f(x) = 2x^2+5$ , evaluate $f'(x)$ using the limit definition above.

The limit definition says

$\begin{eqnarray*} f'(x) &=& \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &=& \lim_{h \to 0} \frac{[2(x+h)^2+5]-[2x^2+5]}{h} \\ &=& \lim_{h \to 0} \frac{\answer{4xh+2h^2}}{h} \\ &=& \lim_{h to 0} \frac{h(\answer{4x+2h})}{h} \\ &=& \lim_{h \to 0} \answer{4x+2h} \\ &=& \answer{4x}. \end{eqnarray*}$

If $f(x) = 2x^2+5$ , find the equation of the tangent line to $y = f(x)$ at $x=-2$ .

In the previous problem, we found $f'(x) = \answer {4x}$ . Plugging in $x=-2$ gives $f'(-2) = \answer {-8}$ . This is the slope of the line. To get a point, we plug in $x=-2$ into $f(x)$ to get the $y$ -coordinate, and this gives $f(-2) = \answer {13}$ . Using point-slope form, we get

$y - \answer {13}= \answer {-8}(x+2).$

If $f(x) = \sqrt {2x}$ , then the limit definition of the derivative tells us that $f'(x) = \answer {\frac {1}{\sqrt {2x}}}$ .

Using the limit definition, we get

$\begin{eqnarray*} f'(x) &=& \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &=& \lim_{h \to 0} \frac{\sqrt{2(x+h)}-\sqrt{2x}}{h} \\ \end{eqnarray*}$

From here we can . In this case, the conjugate of the numerator is

$\answer {\sqrt {2(x+h)}+\sqrt {2x}}.$ Multiplying by this on the top and bottom of the fraction gives

$\begin{eqnarray*} f'(x) &=& \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &=& \lim_{h \to 0} \frac{\sqrt{2(x+h)}-\sqrt{2x}}{h} \\ &=& \lim_{h \to 0} \frac{[\sqrt{2(x+h)}-\sqrt{2x}][\sqrt{2(x+h)}+\sqrt{2x}]}{h[\sqrt{2(x+h)}+\sqrt{2x}]} \\ &=& \lim_{h \to 0} \frac{\answer{2h}}{h[\sqrt{2(x+h)}+\sqrt{2x}]} \\ &=& \lim_{h \to 0} \frac{\answer{2}}{\sqrt{2(x+h)}+\sqrt{2x}} \\ &=& \answer{\frac{1}{\sqrt{2x}}}. \end{eqnarray*}$

The derivative can tell us useful information about a graph. For example, if $f'(a) = 0$ , then we know the tangent line to the graph of $f(x)$ at $x=a$ is . We also know that if $f'(a) >0$ , then the tangent line to the graph of $f(x)$ at $x=a$ has slope. Similarly, if $f'(a) <0$ , then the tangent line to the graph of $f(x)$ at $x=a$ has slope.

Consider the following graph of a function $f(x)$ .

We have $f(2) = \answer {-4}$ and $f(0) = \answer {0}$ .
The limits $\lim _{x \to \infty } f(x) = \answer {\infty }$ and $\lim _{x \to -\infty } f(x) = \answer {-\infty }$ .
The limit $\lim _{x \to 0} f(x) = \answer {0}$ .
Does $f(x)$ have any points of discontinuity?
.
How many horizontal tangent lines does the graph of $f(x)$ have? $\answer {2}$ .
List the points where $f'(x) = 0$ in increasing order: $x = \answer {-2} \quad \text {and} \quad x = \answer {2}.$
The number $f'(0)$ is
zero.
Which of the following is the graph of $f'(x)$ ?

Suppose $f(x) = 3x^2-6x+5$ .

The limit definition of the derivative tells us that $f'(x) = \answer {6x-6}$ .
The graph of $f(x)$ has a horizontal tangent line at $x = \answer {1}$ .
The point on the graph where the tangent line is parallel to $y = 3x+2$ is at $x= \answer {3/2}$ .

For the last part, parallel lines have the same slope. The slope of the tangent line is given by $f'(x)$ , and the slope of the given line is $\answer {3}$ . So set them equal and solve for $x$ .

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