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.

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.

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:

Practice Problem #1

Consider the function $f(x) = \pi.$ Then $f'(x) =$ \[ \begin{array}{lllll} \text{(A) }\pi && \text{(B) }\pi^{-1} && \text{(C) }1 && \text{(D) }0 && \text{(E) none of these} \end{array} \]

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

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.

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*.

*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.[collapse] - 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 $f'(x) =$ \[ \begin{array}{lllll} \text{(A) }0 && \text{(B) }52 && \text{(C) }51 && \text{(D) }-9 && \text{(E) none of these} \end{array} \]

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.*

Hence we have shown that, for integer values of

\[f'(x) = nx^{n-1}\]

The rule also holds for

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

\[ \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.

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 $g'(x) =$
\[ \begin{array}{lllll} \text{(A) }x^3 && \text{(B) }x^5 && \text{(C) }4x^3 && \text{(D) }4 && \text{(E) none of these} \end{array} \]

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 $y'(1).$

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 $\dfrac{1}{x^n} = x^{-n}.$

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

Recall from the Topic “Tangent Line & Curve Slope“:**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:

\begin{align*} y\, -\, y_0 &= m_\text{tangent}(x\, -\, x_0) \\[8px] y\,-\, f(a) &= f'(a)\,(x\, -\, a) \end{align*}

[collapse]

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 \[ \begin{array}{lllll} \text{(A) }0 && \text{(B) }\dfrac{1}{12} && \text{(C) }\dfrac{1}{6} && \text{(D) }\dfrac{1}{2} && \text{(E) none of these} \end{array} \]

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.]

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} [

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 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 $f'(x) = nx^{n-1}.$