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A.1 Derivatives of Constant, Linear, & Power Functions

To begin developing a knowledge of the derivatives of common functions, and the rules we use to find them, let’s start with the most basic of functions: I. Constant; II. Linear; and III. Power Functions. Let’s examine each in turn.

I. Constant Functions

We consider first constant functions: $f(x) = c,$ where $c$ is a constant.

Exploration 1: Compare derivatives of constant functions

Consider the graphs of two constant functions, $f(x) = 20$ and $g(x) = -80.$
To develop the derivative of constant functions, consider two of them: f(x) = 20 and g(x) = -80

To start focus on the rate at which the first function, $y=f(x),$ is changing. (Hint: when you vary the input of the function just a little, how does the function’s output change? What does this tell you about the derivative?) Then imagine the graph of the corresponding derivative function, $f'(x):$ what does this graph look like?

Now think about $\dfrac{dy}{dx}$ for each point on the second graph $y = g(x),$ and imagine the graph of the corresponding derivative function, $g'(x).$ What is its shape? How does it differ from the graph of $f'(x),$ if at all?

When you have your answers in mind, open the area below to use the interactive Desmos calculator to check your reasoning.

Show/Hide Desmos graph of constant function

The upper graph shows the constant function $f(x) = c,$ where initially $c=20.$ You can change the value of the constant function by dragging the red dot up and down the vertical axis.

The lower graph shows the derivative function $f'(x).$ How does it change, if at all, given different values for c?

Graph of f(x) = c versus x


$\color{green}{\Big\Downarrow}$Graph of f ‘(x) versus x $\color{green}{\Big\Downarrow}$

Note that the derivative of the constant function $f(x)=c$ is always zero, $f'(x) =0,$ no matter what the value of c is.

How would you explain this result to a friend who knows a little bit of Calculus?

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As Exploration 1 demonstrates,
\[ \bbox[10px,border:2px solid blue]{
\begin{align*}
\textbf{Constant Functions:} & \\[4px] \text{If $f(x) = c,$ where } c &\text{ is a constant, then $f'(x) = 0.$}
\end{align*} }\] Here are three ways to make sense of this result:

  • The derivative is the rate of change of the function. Since the function $f(x) = c$ never changes, its rate of change is zero.
  • The value of the derivative at any point is the slope of the tangent line to the curve $y =f(x)$ at that point. The graph of a constant function is a horizontal line, which has zero slope; hence the derivative is zero everywhere.
  • We can obtain this result using the definition of the derivative:
  • We have $f(x) = c$, so $f(x+h) = c.$ Hence
    \begin{align*}
    f'(x) &= \lim_{h \to 0}\dfrac{f(x+h) – f(x)}{h} \\[8px] &= \lim_{h \to 0}\dfrac{c\, -\, c}{h} \\[8px] &= \lim_{h \to 0}\dfrac{0}{h} = 0 \quad \cmark
    \end{align*}


Let’s do one quick Practice Problem to help solidify this result:

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II. Linear Functions

Let’s consider next linear functions, which we’ll write in the familiar form $f(x) = mx + b.$

Exploration 2: Compare derivatives of linear functions

To develop the derivative of linear functions, consider two of them: f1 and f2 have the same slope, but different y-intercepts: f1(x) = 2x + 3 and f2(x) = 2x -3
Examine the graphs of the two linear functions $f_{1}(x)$ and $f_{2}(x).$ The two lines have the same slope, but $f_1$ has a y-intercept of $b_1 = +3$ while $f_2$ has a y-intercept of $b_2 = -3.$ Imagine the graphs of the corresponding derivative functions $f_{1}'(x)$ and $f_{2}'(x).$ What are their shapes? How do they differ, if at all?
To develop the derivative of linear functions, consider another two of them: f1 and f3 have the same y-intercept, but different slopes: f1(x) = 2x+3 and f3(x) = -2x+3
Next, examine the graphs of the two linear functions $f_1(x)$ and $f_3(x).$ They have the same y-intercept, but $f_1$ has a slope of $m_1 = +2$ while $f_3$ has a slope of $m_3 = -2.$ Imagine the graphs of the corresponding derivative functions, $f_{1}'(x)$ and $f_{3}'(x).$ What are their shapes? How do they differ, if at all?

When you have your answers in mind, open the are below to use an interactive Desmos calculator to check your reasoning.

Show/Hide Desmos graph of linear function

The upper graph below initially displays the function $f(x) = 2x + 3.$

  • To adjust the y-intercept value, drag the red dot up and down along the y-axis.
  • To adjust the slope m of the line, drag the second, purple dot up and down.


As usual, the lower graph displays the corresponding derivative function $f'(x).$ How does the derivative function change as you vary b? m?

Graph of $f(x) = mx + b$ versus x


$\color{green}{\Big\Downarrow}$Graph of f ‘(x) versus x $\color{green}{\Big\Downarrow}$

Note that the derivative of the linear function $f(x) = mx + b$ is always $f'(x) = m,$ regardless of the value of b. Does this behavior match the predictions you made above? How would you explain this result to a friend who knows a little bit of Calculus?

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As Exploration 2 demonstrates,
\[ \bbox[10px,border:2px solid blue]{
\begin{align*}
\textbf{Linear Functions:} & \\[4px] \text{If }f(x) = mx+b, &\text{ then $f'(x) = \text{slope} =m.$}
\end{align*}
}\] Here are three ways to make sense of this result:

  • The derivative is the rate of change of the function. By definition, a linear function changes at the constant rate given by its slope, m. The function’s y-intercept doesn’t matter; that value just tells us one point on the line, $(0,b),$ and does nothing to affect the rate of change.

    We can visualize this result more clearly by tying it back to Leibniz notation, as shown in the following graph.

    Show/Hide graph of line with adjustable dx and dy.

    The graph below shows the line $y = mx + b.$ Use the sliders to vary the values of b and m.


    Slope: m =

    y-intercept: b =
    The black dot represents a particular point, $(a, f(a)),$ on the function. To change the value of a, drag the black dot on the line. To change the size of dx, drag the red dot.


    Currently a = .   
    At this point (and at every point on the line), dy = dx. That is, thinking of the derivative as a “measure of the function’s reactivity,” when we vary the input by the small amount dx, the function reacts by changing its output dy by times as much.

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  • The value of the derivative at any point is the slope of the tangent line to the curve at that point. For a linear function, the tangent line to the “curve” is the line itself. Hence the derivative is simply equal to the line’s slope.
  • We can obtain this result using the definition of the derivative:
    We have $f(x) = mx+b$, so
    \begin{align*}
    f(x+h) &= m(x+h) + b \\[8px] &= mx + mh + b
    \end{align*}
    Hence
    \begin{align*}
    f'(x) &= \lim_{h \to 0}\dfrac{f(x+h) – f(x)}{h} \\[8px] &= \lim_{h \to 0}\dfrac{(mx + mh + b) – (mx + b)}{h} \\[8px] &= \lim_{h \to 0}\dfrac{mh}{h} \\[8px] &= \lim_{h \to 0}m = m \quad \cmark
    \end{align*}


Let’s again do one quick Practice Problem to help ingrain this result:

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III. Power Functions

Let’s consider now functions of the form $f(x) = x^n,$ such as $x^2,$ $x^3,$ $x^{1/2} = \sqrt{x},$ and $x^{-1} = 1/x.$ We’ve actually computed each of those particular derivative functions earlier. [We’ll add these links once we make Chapter 3 available.] We’ve provided those results below, in both graphical and equation form.
To develop the derivative of power functions, consider the graphs of the following functions and their derivatives. a: f(x) = x^2, f'(x) = 2x. b: f(x) = x^3, f'(x) = 3x^2. c: f(x) = sqrt(x) = x^(1/2), f'(x) = (1/2)x^(-1/2). d: f(x) = 1/x = x^(-1), f'(x) = -x^(-2).
Do you notice a pattern in how the derivative functions are related to the original functions? Look at this table for more examples, and the last entry, which shows how to summarize the pattern:
\[ \begin{array}{c|c|c|c|c|c|c|c|c}
f(x): & x^{-3} & x^{-2} & x^{-1} & x^0 & x^1 & x^2 & x^3 & x^n\\
\Downarrow & \Downarrow & \Downarrow & \Downarrow & \Downarrow &\Downarrow & \Downarrow & \Downarrow & \Downarrow\\
f'(x): & -3x^{-4} & -2 x^{-3} & -x^{-2} & 0 & 1 \left(x^0 \right) & 2x^{(1)} & 3x^2 & nx^{n-1}
\end{array} \] Of course the fact that we can recognize a pattern from a small set of examples is not sufficient for us to conclude that the pattern always holds; instead, we must prove that we have in fact developed a rule we can trust, or determine when the rule holds and when it doesn’t.

Using the tools we already have, we can in fact prove that the pattern holds for any positive integer value of n:

Proof that if $f(x) = x^n,$ then $f'(x) = nx^{n-1}$ for positive integer values of n.

Show/Hide Proof

It’s easiest to prove that the pattern holds by returning to the first form of the Definition of the Derivative:
\[f'(a) = \lim_{x \to a}\frac{f(x) – f(a)}{x-a}\] We have $f(x) = x^n.$ Hence
\[f'(a) = \lim_{x \to a}\frac{x^n – a^n}{x-a}\] It turns out that for positive integer values of n, we can rewrite the numerator by using the identity:
\[x^n – a^n = (x-a)\overbrace{\left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \ldots + x^2a^{n-3} + xa^{n-2} + a^{n-1}\right)}^{n \text{ terms}} \]
Show/Hide verification of the preceding identity

To verify the identity, we simply multiply out the right hand side of the equation and then see how most of the terms cancel:
\begin{align*}
x^n – a^n &= (x-a)\left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \ldots + x^2a^{n-3} + xa^{n-2} + a^{n-1}\right) \\[8px] &= x \left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \ldots + x^2a^{n-3} + xa^{n-2} + a^{n-1}\right) \\[8px] & \qquad – a \left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \ldots + x^2a^{n-3} + xa^{n-2} + a^{n-1}\right) \\[8px] &= \left(x^n + x^{n-1}a + x^{n-2}a^2 + \ldots + x^3a^{n-3} + x^2a^{n-2} + xa^{n-1}\right) \\[8px] & \qquad – \left(x^{n-1}a + x^{n-2}a^2 + x^{n-3}a^3 + \ldots + x^2a^{n-2} + xa^{n-1} + a^n\right) \\[8px] &= x^n + \left(x^{n-1}a + x^{n-2}a^2 + \ldots + x^3a^{n-3} + x^2a^{n-2} + xa^{n-1}\right) \\[8px] & \qquad – \left(x^{n-1}a + x^{n-2}a^2 + x^{n-3}a^3 + \ldots + x^2a^{n-2} + xa^{n-1} \right) – a^n\\[8px] &= x^n – a^n \quad \cmark
\end{align*}
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Let’s make that identity-substitution and then continue to find the derivative:
\begin{align*}
f'(a) &= \lim_{x \to a}\frac{x^n – a^n}{x-a} \\[8px] &= \lim_{x \to a}\frac{(x-a)\overbrace{\left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \ldots + x^2a^{n-3} + xa^{n-2} + a^{n-1}\right)}^{n \text{ terms}}}{x-a} \\[8px] &= \lim_{x \to a}\frac{\cancel{(x-a)}\left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \ldots + x^2a^{n-3} + xa^{n-2} + a^{n-1}\right)}{\cancel{x-a}} \\[8px] &= \lim_{x \to a} \left(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \ldots + x^2a^{n-3} + xa^{n-2} + a^{n-1}\right) \\[8px] &= a^{n-1} + a^{n-2}a + a^{n-3}a^2 + \ldots + a^2a^{n-3} + a a^{n-2} + a^{n-1} \\[8px] &= \overbrace{a^{n-1} + a^{n-1} + a^{n-1} + \ldots + a^{n-1} + a^{n-1} + a^{n-1}}^{n \text{ terms}} \\[8px] &= n a^{n-1} \quad \cmark
\end{align*}
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Hence we have shown that, for integer values of n, the derivative of the function $f(x) = x^n$ is
\[f'(x) = nx^{n-1}\]

The rule also holds for negative integer values of n, but we must wait until later in the course when we have more tools to be able to prove that this is so. [Link to proof to be added.]

 

Furthermore, later we’ll also be able to show [Link to be added] that the rule also holds for fractional values of n, such as
\[ \begin{array}{c|c|c|c|c|c|c|c|c|c|}
f(x): & x^{-2/3} & x^{-1/2} & x^{1/2} & x^{2/3} & x^n\\
\Downarrow & \Downarrow & \Downarrow & \Downarrow & \Downarrow & \Downarrow\\
f'(x): & \frac{-2}{3}x^{-5/3} & \frac{-1}{2} x^{-3/2} &\frac{1}{2}x^{-1/2} & \frac{2}{3}x^{-1/3} & nx^{n-1}
\end{array} \]

But wait, there’s more [Link again to be added]: the rule also holds for decimal values of n that can be expressed as a fraction (e.g., $0.4 = \frac{4}{10}$) – meaning all rational values of n – such as:
\[ \begin{array}{c|c|c|c|c|c|c|c|c|c|}
f(x): & x^{-0.4} & x^{-0.2} & x^{0.2} & x^{0.4} & x^n\\
\Downarrow & \Downarrow & \Downarrow & \Downarrow & \Downarrow & \Downarrow \\
f'(x): & -0.4x^{-1.4} & -0.2x^{-1.2} & 0.2x^{-0.8} & 0.4x^{-0.6} & nx^{n-1}
\end{array} \] And finally, the rule also holds [Final link to be added] for non-rational values of n (values that cannot be expressed as a fraction) such as $\pi,$ e, and $\sqrt{2}$:
\begin{array}{c|c|c|c|c|c|c|c|c|c|}
f(x): & x^{\pi} & x^e & x^\sqrt{2} & x^n\\
\Downarrow & \Downarrow & \Downarrow & \Downarrow & \Downarrow \\
f'(x): & \pi x^{\pi-1} & ex^{e-1} & \sqrt{2}x^{\sqrt{2} – 1} & nx^{n-1}
\end{array}

Hence, the rule holds for all real values of n. Although we haven’t proven it fully yet, from this point onward we will use the general rule, known as the Power Rule:
\[ \bbox[10px,border:2px solid blue]{ \begin{align*}
\textbf{Power Rule:} & \\[4px] \text{If }f(x) &= x^n, \text{ then } f'(x) = nx^{n-1} \\[4px] \text{for } &\textit{any} \text{ real value of }n.
\end{align*} }\] You can see the graph of $f(x) = x^n$ and its derivative $f'(x) = nx^{n-1}$ in the following Exploration.

Exploration 3: Graphs of $f(x) = x^n$ and its derivative $f'(x) = nx^{n-1}$

The upper graph below shows the function $f(x) = x^n.$ Initially $n=3,$ so $f(x) = x^3,$ but you can change the value using the slider to any value such that $-6 \le n \le 6.$

The graph underneath shows the derivative function $f(x) = nx^{n-1}.$ It will automatically update as you change n in the upper graph.

Use this slider to set the value of n: n =
Graph of $f(x) = x^n$ versus x


$\color{green}{\Big\Downarrow}$Graph of f ‘(x) versus x $\color{green}{\Big\Downarrow}$
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Let’s revisit a problem from Chapter 1, where we found the linear approximation to $\sqrt{16.2}.$ At that point, we had to simply tell you the rate at which the function changes at $x = 16,$ but now you can find that rate for yourself!

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On the next screen we’ll add to our toolkit of derivatives we can quickly compute, by examining the exponential function $a^x.$


Of course if you have questions or thoughts about anything on this screen, please let us know on the Forum!


The Upshot

    The derivatives of the following functions are:

  1. Constant Function: If $f(x) = c,$ where c is a constant, then $f'(x) = 0.$
  2. Linear Function: If $f(x) = mx+b,$ then $f'(x) = \text{slope} =m.$
  3. Power Rule: If $f(x) = x^n,$ then $f'(x) = nx^{n-1}.$