Let’s use Desmos to explore derivatives of the most basic of functions: I. Constant; II. Linear; and III. Power Functions. We’ll develop a simple rule to quickly determine each, and the of course practice using each rule to quickly find the derivative.
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 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:
Let’s do one quick Practice Problem to help solidify this result:
Practice Problem #1
Consider the function $f(x) = \pi.$ Then $f'(x) =$ \[ \begin{array}{lllll} \text{(A) }\pi && \text{(B) }\pi^{-1} && \text{(C) }1 && \text{(D) }0 && \text{(E) none of these} \end{array} \]
Oops: recall that the derivative of a constant–any fixed number–is 0. Always. Please answer again.
Oops: recall that the derivative of a constant–any fixed number–is 0. Always. Please answer again.
Oops: recall that the derivative of a constant–any fixed number–is 0. Always. Please answer again.
Great – you got it!
Oops: recall that the derivative of a constant–any fixed number–is 0. Always. Please answer again.
Show/Hide Solution
$\pi$ is simply a number, a constant. Hence f(x) = constant, and so \[ f'(x) = 0 \implies \text{(D)} \quad \cmark \]
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[hide solution]
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
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? 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:
Practice Problem #2
Consider the function $f(x) = 52x - 9.$ Then $f'(x) =$ \[ \begin{array}{lllll} \text{(A) }0 && \text{(B) }52 && \text{(C) }51 && \text{(D) }-9 && \text{(E) none of these} \end{array} \]
Oops: recall that the derivative of a linear function $f(x) = mx + b$ is $f'(x) = m,$ the slope of the line. Please answer again.
Correct!
Oops: recall that the derivative of a linear function $f(x) = mx + b$ is $f'(x) = m,$ the slope of the line. Please answer again.
Oops: recall that the derivative of a linear function $f(x) = mx + b$ is $f'(x) = m,$ the slope of the line. Please answer again.
Oops: recall that the derivative of a linear function $f(x) = mx + b$ is $f'(x) = m,$ the slope of the line. Please answer again.
Show/Hide Solution
The derivative of a linear function $f(x) = mx + b$ is $f'(x) = m,$ the slope of the line. Hence \[f'(x) = 52 \implies \text{(B)} \quad \cmark \]
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[hide solution]
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. 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*}
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 $\color{green}{\Big\Downarrow}$
The first three Practice Problems below are straightforward applications of the Power Rule. The next few provide practice at “find the derivative” as the first step in writing the equation for a tangent line to a curve, one of the most common homework (and exam!) questions you’ll encounter.
Practice Problem #3
Consider the function $g(x) = {x^4}.$ Then $g'(x) =$
\[ \begin{array}{lllll} \text{(A) }x^3 && \text{(B) }x^5 && \text{(C) }4x^3 && \text{(D) }4 && \text{(E) none of these} \end{array} \]
Recall that the derivative of $x^n$ is $nx^{n-1}.$ Here $n=4.$ Please answer again.
Recall that the derivative of $x^n$ is $nx^{n-1}.$ Here $n=4.$ Please answer again.
You got it!
Recall that the derivative of $x^n$ is $nx^{n-1}.$ Here $n=4.$ Please answer again.
Recall that the derivative of $x^n$ is $nx^{n-1}.$ Here $n=4.$ Please answer again.
Show/Hide Solution
Our function $g(x) = x^4$ is in the form of $x^n,$ with $n = 4.$ So applying the Power Rule $\left(g'(x) = nx^{n-1} \right)$ gives us
\begin{align*}
g'(x) &= 4x^{(4-1)} \\[8px]
&= 4x^3 \implies \text{(C)} \quad \cmark
\end{align*}
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Practice Problem #4
Consider the function $y(t) = t^{99}.$ Then $y'(1) =$
\[ \begin{array}{lllll} \text{(A) }0 && \text{(B) }99 && \text{(C) }98 && \text{(D) }1 && \text{(E) none of these} \end{array} \] Hint: First find the derivative $y'(t).$ Then substitute $t=1$ to find $y'(1).$
Recall that the derivative of $t^n$ is $nt^{n-1}.$ Here $n=99,$ so $y'(t) = 99t^{98}.$ Then $y'(1) = ...$
Excellent.
Recall that the derivative of $t^n$ is $nt^{n-1}.$ Here $n=99,$ so $y'(t) = 99t^{98}.$ Then $y'(1) = ...$
Recall that the derivative of $t^n$ is $nt^{n-1}.$ Here $n=99,$ so $y'(t) = 99t^{98}.$ Then $y'(1) = ...$
Recall that the derivative of $t^n$ is $nt^{n-1}.$ Here $n=99,$ so $y'(t) = 99t^{98}.$ Then $y'(1) = ...$
Show/Hide Solution
Our function $y(t) = t^{99}$ is in the form of $t^n,$ with $n = 99.$ So applying the Power Rule $\left(y'(t) = nt^{n-1} \right)$ gives us
\begin{align*}
y'(t) &= 99t^{(99-1)} \\[8px]
&= 99t^{98}
\end{align*}
Then substituting t = 1 gives:
\begin{align*}
y'(t) &= 99 \cdot (1)^{98} \\[8px]
&= 99 \cdot (1) = 99 \implies \text{(B)} \quad \cmark
\end{align*}
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[hide solution]
Practice Problem #5
Consider the function $h(x) = \dfrac{1}{x^3}.$ Then $h'(x) =$ \[ \begin{array}{lllll} \text{(A) }\dfrac{3}{x^2} && \text{(B) } \dfrac{-1}{3x^4} && \text{(C) }-\dfrac{3}{x^4} && \text{(D) }\dfrac{1}{3x^2} && \text{(E) none of these} \end{array} \] Hint: Recall that $\dfrac{1}{x^n} = x^{-n}.$
Show/Hide Solution
First, recall that \[\frac{1}{x^3} = x^{-3} \] Hence we have $h(x) = x^{-3},$ which is in the form of $x^n$ where $n = -3.$ So applying the Power Rule $\left(h'(x) = nx^{n-1} \right)$ gives us \begin{align*} h'(x) &= -3x^{-4} \\[8px] &= -\frac{3}{x^4} \implies \text{ (C)} \quad \cmark \end{align*}
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Practice Problem #6
An equation for the tangent line to the curve $y = x^5$ at $x = 1$ is \[ \begin{array}{ll} \text{(A) }y = 5x && \text{(B) }y - 1 = 5(x-1) \end{array} \] \[ \begin{array}{lll} \text{(C) } y -1 = 5x - 1 && \text{(D) } y+1 = 5(x+1) && \text{(E) none of these} \end{array} \]
Show/Hide steps to find equation for the tangent line to a curve
Find the $y$-value of the point of interest: $y_0 = f(a).$
Find the value of the derivative at the point of interest, $f'(a).$
Use these two pieces of information, $m_\text{tangent} = f'(a)$, and the point of interest $(x_0, y_0),$ to write the equation of the tangent line in Point-Slope form:
Step 1. Find the $y$-value of the point of interest: $y_0 = f(a).$ We’re interested in the point $x=1,$ so have $x_0 = a = 1.$ Since $y = f(x) = x^5,$ \[y_0 = f(1) = (1)^5 = 1 \] Hence we’re looking for the tangent line to the curve at the point $(1, 1). \,\blacktriangleleft$ Step 2. Find the value of the derivative at the point of interest, $f'(a).$ Since $f(x) = x^5,$ from the Power Rule we know \[f'(x) = 5x^4 \] Hence \[f'(1) = 5(1)^4 = 5 \, \blacktriangleleft\] Step 3. Use these two pieces of information to write the equation of the tangent line. (See the hint above for more information about this step.) The equation of the tangent line to a curve is given by \[y\, – \,f(a) = f'(a)\,(x\, -\, a) \] Using the information we found above, then, we have \[y\, -\, 1 = 5(x-1) \implies \text{( B)} \quad \cmark \]
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[hide solution]
Practice Problem #7
The value of the slope of the tangent line to the curve $y = x^{1/3}$ at the point $(8, 2)$ is \[ \begin{array}{lllll} \text{(A) }0 && \text{(B) }\dfrac{1}{12} && \text{(C) }\dfrac{1}{6} && \text{(D) }\dfrac{1}{2} && \text{(E) none of these} \end{array} \]
Show/Hide Solution
Recall that the slope of the tangent line equals the derivative of the function at that point. Hence, given $f(x) = x^{1/3}$ we need to find $f'(8).$ We have $f(x) = x^n,$ where $n = \dfrac{1}{3}.$ Then by the Power Rule $f'(x) = nx^{n-1}$: \[f'(x) = \frac{1}{3}x^{-2/3} \] and \begin{align*} f'(8) &= \frac{1}{3}(8)^{-2/3} \\[8px] &= \frac{1}{3}\left(8^{1/3} \right)^{-2} \\[8px] &= \frac{1}{3}(2)^{-2} \\[8px] &= \frac{1}{3} \cdot \frac{1}{4} \\[8px] &= \frac{1}{12} \implies \text{(B)} \quad \cmark \end{align*}
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[hide solution]
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!
Practice Problem #8
Consider the square-root function $f(x) = \sqrt{x}.$ You know that $f(16) = \sqrt{16} = 4.$ Using our linear approximation method, the approximate value of $\sqrt{16.2}$ is
\begin{array}{lllll} \text{(A) }4.02 && \text{(B) }4.0249 && \text{(C) }4.025 && \text{(D) }4.0125 && \text{(E) none of these} \end{array}
[Hint: First, find the rate at which the function $f(x) = \sqrt{x}$ changes at $x = 16.$ Then if you need to review linear approximations, go here.]
Excellent!!
Show/Hide Solution
Our solution requires the rate at which the function changes at $x = 16,$ which is $f'(16).$ We thus need the Power Rule, as applied to $f(x) = \sqrt{x} = x^{1/2}$:
\begin{align*}
f(x) &= \sqrt{x} = x^{1/2} \\[8px]
f'(x) &= \dfrac{1}{2}x^{-1/2} \\[8px]
&= \dfrac{1}{2}\dfrac{1}{\sqrt{x}} \\[8px]
f'(16) &= \dfrac{1}{2}\dfrac{1}{\sqrt{16}} \\[8px]
&= \dfrac{1}{8} = 0.125 \; \blacktriangleleft
\end{align*}
That’s the rate we simply had to provide back in Chapter 1, but that you can now find for yourself. Now knowing this rate, the solution is identical to that from Chapter 1:
\begin{align*}
\overbrace{f(16+dx)}^\text{$f$ value at $16+dx$} &\approx \overbrace{f(16)}^{\text{$f$ value at }16} + \overbrace{\text{ small change }df}^{\text{(rate at $x=16$)} * \, (dx)} \\[8px]
f(16 + 0.2) &\approx 4 + (0.125)(0.2) \\[8px]
&= 4 + 0.025 \\[8px]
&= 4.025 \implies \text{ (C)} \quad \cmark
\end{align*}
For reference, the actual numeric value to 6 decimal places is $\sqrt{16.2} = 4.02492236.$
You can use the interactive Desmos graph below to explore this result graphically. Once again, visuals for dx and df will appear once you have zoomed-in sufficiently, which you must do to start. You can also try out different values of dx by using the slider beneath the graph.
Graph of $f(x) = \sqrt{x}$ versus x
Use the slider to change the value of dx:
Currently dx = 0.2
The linear approximation for the current value of dx is:
\begin{align*}
\overbrace{f(16+dx)}^\text{$f$ value at $16+dx$} &\approx \overbrace{f(16)}^{\text{$f$ value at }16} + \overbrace{\text{ small change }df}^{\text{(rate at $x=16$)} * \, (dx)} \\[8px]
f(16+0.2) &\approx 4+0.125(0.2) \\[8px]
&=4+0.025= 4.025
\end{align*}
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[hide solution]
On the next screen we’ll add to our toolkit of derivatives we can quickly compute, by examining the exponential function $a^x.$
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The Upshot
The derivatives of the following functions are:
Constant Function: If $f(x) = c,$ where c is a constant, then $f'(x) = 0.$
Linear Function: If $f(x) = mx+b,$ then $f'(x) = \text{slope} =m.$
Power Rule: If $f(x) = x^n,$ then $f'(x) = nx^{n-1}.$
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