Calculus Workbook: The Derivative as a Function

Section 2.2 The Derivative as a Function

Objectives

State the limit definition of the derivative function and use it to compute derivative functions
Sketch and interpret slope graphs
Use and interpret common derivative notations for derivatives of any order
Describe what it means for a function to be differentiable, determine where a function is differentiable, identify and give examples of the ways in which a function fails to be differentiable
Describe the relationship between continuity and differentiability
Interpret first, second, and third derivatives in the context of motion.

Subsection 2.2.1 Before Class

https://mymedia.ou.edu/media/2.2-1/1_fb43p7ws

Figure 12. Pre-Class Video 1

https://mymedia.ou.edu/media/2.2-2/1_g1ws2wmb

Figure 13. Pre-Class Video 2

Subsubsection 2.2.1.1 The Derivative as a Function

Example 2.2.1.

Use either method of the previous section to find the derivative of $f(x) = x^2 + 1$ at $(1,2)\text{.}$
Find the derivative of $f(x)$ at $(2,5)\text{.}$

Solution.

$\displaystyle f'(1) = 2$
$\displaystyle f'(2) = 4$

Definition 2.2.2. The Derivative (as a Function).

The derivative of a function $f(x)$ is given as

\begin{equation*} \lim_{h\to 0} \dfrac{f(x+h) - f(x)}{h} \end{equation*}

where $h$ is a small change in $x\text{.}$

Example 2.2.3.

Use the previous definition to compute $f'(x)\text{,}$ if $f(x) = x^2 + 1\text{.}$
Find $f'(1)$ and $f'(2)\text{.}$

Solution.

$\displaystyle f'(x) = 2x$
$f'(1) = 2$ and $f'(2) = 4$

Subsubsection 2.2.1.2 Differentiability & Non-Differentiability

Definition 2.2.4. Differentiability.

A function $f$ is differentiable at $a$ if $f'(a)$ exists. It is differentiable on an open interval $(a,b)$ if it is differentiable at every number in the interval.

A Note.

Here is the formal definition of differentiability: a function $f(x)$ is differentiable at $a$ if

\begin{equation*} \lim_{h\to 0^-} \dfrac{f(x+h)-f(x)}{h} = \lim_{h\to 0^+} \dfrac{f(x+h)-f(x)}{h} \end{equation*}

Example 2.2.5.

Where is $f(x) = |x|$ differentiable?

Solution.

Everywhere but $x=0$

Example 2.2.6.

Where is $g(x) = [\!x]\!]$ differentiable?

Solution.

At non-integer values

Question 2.2.7.

Rewrite the definition of continuity. What relationship(s) do you see between continuity and differentiability?

Solution.

Continuity requires that $\ds \lim_{x\to a} f(x) = f(a)\text{;}$ for differentiability, we require that $\ds \lim_{x\to a} \dfrac{f(x)-f(a)}{x-a}\text{.}$ Rearrangement of the differentiability condition should give something similar the condition for continuity.

Differentiability vs. Continuity.

If a function $f$ is differentiable, then it is continuous; the reverse is not true.

There are several characteristics which indicate that a graph is not differentiable at a point:

A sharp corner
A discontinuity
A vertical tangent line

Subsubsection 2.2.1.3 Other Notations

\begin{equation*} f'(x) = y' = \dfrac{dy}{dx} = \dfrac{df}{dx} = \dfrac{d}{dx}[f(x)] = Df(x) = D_xf(x) \end{equation*}

When we want to evaluate a derivative, we often use a vertical bar to communicate that we are evaluating. If $f(x) = x^2\text{,}$ then $f'(x) = 2x\text{,}$ and $f'(1) = 2\text{;}$ we could also notate this by writing

\begin{equation*} \dfrac{dy}{dx}\bigg\rvert_{x = 1} = 2 \end{equation*}

This is read as "evaluating $\dfrac{dy}{dx}$ at $x=1$ gives 2" or "$\dfrac{dy}{dx}$ evaluated at $x=1$ is 2"

Subsection 2.2.2 Pre-Class Activities

Example 2.2.8.

What are some similarities and differences between the definitions of the derivative in Section 2.1 and the one in this section?

Solution.

Answers vary

Example 2.2.9.

For any linear function $f(x) = ax + b\text{,}$ $f'(x) = a\text{.}$ Try showing this with the formal definition of the derivative. Without computation, why would this be true?

Solution.

The tangent line to a linear function is the linear function itself, so the slope of the tangent line is the same as the slope of the line.

Example 2.2.10.

Let $P$ represent the percentage of a city’s electrical power that is produced by solar panels, $t$ years after January 1, 2015.

What does $\dfrac{dP}{dt}$ represent in this context? Include the units.
Convert the statement $\dfrac{dP}{dt}\bigg\rvert_{t = 3} = 2.1$ to an English sentence.
Convert the English sentence into derivative notation: Five years after 2015, the percentage of a city’s electrical power produced by solar panels was decreasing by 0.7 percent per year.

Solution.

The rate of change of percentage of electrical power produced by solar panels, in percent per year
The percentage of the city’s power produced by solar is increasing by 2.1% per year, 3 years after 1/1/15.
$\displaystyle \dfrac{dP}{dt}\bigg\rvert_{t=5} = -0.7$

Example 2.2.11.

The graph of a function $f$ is given. State the input values at which $f$ is not differentiable.

Solution.

$f$ is not differentiable at $x=-4,x=0\text{,}$ and $x=2.5$

Subsection 2.2.3 In Class

Subsubsection 2.2.3.1 Derivatives

Example 2.2.12.

Find a formula for $g'(x)\text{,}$ if $g(x) = x^3 + 1$

Solution.

$g'(x) = 3x^2$

Even in this relatively benign example, the algebra got complicated. Here are three algebraic reminders:

$(x+a)^n\neq x^n + a^n\text{.}$ Please, do not make this mistake!
$\displaystyle (x+a)^2 = x^2 + 2ax + a^2$
$\displaystyle (x+a)^3 = x^3 + 3x^2a + 3xa^2 + a^3$

Four-Step Method.

Write (and expand) $f(x+h)$
Subtract $f(x)$ from $f(x+h)$
Divide #2 by $h$
Take the limit as $h\to 0$

Example 2.2.13.

Use the four-step method to algebraically calculate $f'(x)\text{,}$ if $f(x) = x^2 - 3x -1\text{.}$

Solution.

\begin{align*} f(x+h) &= (x+h)^2 - 3(x+h) -1\\ &= x^2+2xh+h^2-3x-3h-1\\ f(x+h)-f(x) &= (x^2+2xh+h^2-3x-3h-1)-(x^2-3x-1)\\ &= 2xh+h^2-3h\\ \dfrac{f(x+h)-f(x)}{h} &= 2x+h-3\\ \ds \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} &= \ds \lim_{h\to 0} 2x + h -3\\ &= 2x-3 \end{align*}

Example 2.2.14.

Find $h'(x)\text{,}$ if $h(x) = \sqrt{x}$

Solution.

$h'(x) = \dfrac{1}{2\sqrt{x}}\text{,}$ or $h'(x) = \dfrac{1}{2}x^{-1/2}$

The previous example gives rise to one of several common tips/tricks when algebraically calculating derivatives:

Combine fractions
Multiply by a conjugate

Example 2.2.15.

Let $q(t) = \sqrt{1+t}\text{.}$ Show that $\dfrac{dq}{dt} = \dfrac{1}{2}(1+t)^{-1/2}\text{.}$

Solution.

\begin{align*} q(t+h) &= \sqrt{1+t+h} \\ q(t+h)-q(t) &= \sqrt{1+t+h}-\sqrt{1+t}\\ \dfrac{q(t+h)-q(t)}{h} &= \dfrac{\sqrt{1+t+h}-\sqrt{1+t}}{h}\cdot \dfrac{\sqrt{1+t+h}+\sqrt{1+t}}{\sqrt{1+t+h}+\sqrt{1+t}}\\ &= \dfrac{(1+t+h)-(1+t)}{h(\sqrt{1+t+h}+\sqrt{1+t})}\\ &= \dfrac{1}{\sqrt{1+t+h}+\sqrt{1+t}}\\ \ds \lim_{h\to 0} \dfrac{q(t+h)-q(t)}{h} &= \ds \lim_{h\to 0} \dfrac{1}{\sqrt{1+t+h}+\sqrt{1+t}}\\ &= \dfrac{1}{2\sqrt{1+t}}\\ &= \dfrac{1}{2}(1+t)^{-1/2} \end{align*}

Example 2.2.16.

Let $k(x) = \dfrac{3+x}{x-2}\text{.}$ Find $k'(x)\text{.}$

Solution.

$-\dfrac{5}{(x-2)^2}$

Subsubsection 2.2.3.2 Slope Graphs

Example 2.2.17.

Use the graph of $f$ below to sketch the graph of $f'\text{.}$

$The graph of the function $.5*x^3-3.5*x^2+6*x+1.3\text{,}$ on the interval $[0,5]\text{.}$ There are dotted lines drawn from the maximum at $x=1.13$ to the $x-$axis, from another point $x=2.33$ to the $x-$axis, and from the minimum $x=3.53$ to the $x-$axis.$

$This is a blank graph for sketching the derivative, $f'(x)\text{,}$ on the same interval/scale as $f(x)\text{.}$$

Solution.

$This is the graph of the derivative of the function from the example.$

Example 2.2.18.

Match the function in (a) - (d) with its derivative graph in (I) - (IV). Give reasons for each choice.

Solution.

Example 2.2.19.

The graph of $g(t)$ is given. Sketch a possible graph of $g'(t)\text{.}$

$This is the graph of $(0.25x)^4-(0.25x)^2\text{.}$ There are two minima around 1 and $-1\text{,}$ and a maximum at 0.$

Solution.

$This is the graph of the derivative in the example. There is a maximum around $-1$ and a minimum around x=1, with an inflection point at 0.$

Example 2.2.20.

Consider the function $y=a\cos x\text{.}$ Find $\dfrac{df}{dx}$ and $\dfrac{df}{da}$

Solution.

$\dfrac{df}{dx} = -a\sin x$ and $\dfrac{df}{da} = \cos x$

Subsubsection 2.2.3.3 Higher Derivatives

In physics, higher-order derivatives are very important. If $s(t)$ is a position function, then the first derivative $s'(t)$ is called the velocity, often written as $v(t)\text{.}$ The second derivative of position (so, the first derivative of velocity) is called acceleration, often written as $a(t)\text{.}$ The third derivative also has a special name, called the jerk, $j(t)\text{.}$

Notation for Higher-Order Derivatives.

Second derivative: $f''(x)\text{,}$ $\dfrac{d^2y}{dx^2}$
Higher Derivatives: $f^{(n)}(x)\text{,}$ $\dfrac{d^ny}{dx^n}$

Example 2.2.21.

If $f(x) = x^3-x\text{,}$ find $f''(x)\text{,}$ $\dfrac{d^3f}{dx^3}\text{,}$ and $f^{(4)}(x)\text{.}$

Solution.

$f''(x) =3x^2-1\text{,}$ $\dfrac{d^3f}{dx^3} = 6x\text{,}$ and $f^{(4)}(x) = 6$

Subsection 2.2.4 After Class Activities

Example 2.2.22.

Find the first two derivatives of $a(x) = \sqrt{9-x}\text{.}$ Write your answer using both notations for the derivative.

Solution.

$a'(x) = \dfrac{da}{dx} = -\dfrac{1}{2}(9-x)^{-1/2}$ and $a''(x) = \dfrac{d^2a}{dx^2} = -\dfrac{1}{4}(9-x)^{-3/2}$

Example 2.2.23.

The graph of $f$ is given. Sketch a possible graph for $f'\text{.}$

Solution.

Answers vary