On this screen we're going to use Desmos to examine the derivative of exponential functions like 2𝑥,3.25𝑥, and such.
Exploration of the derivative of 𝑎𝑥
Let's use Desmos to examine the derivative of 𝑎𝑥 for various values of 𝑎.
Exploration 1
[End Exploration 1]
Let's see how the discovery we made in Exploration 1 follows from the definition of the derivative as applied to the exponential function 𝑓(𝑥)=𝑎𝑥. Recall the definition of the derivative:
𝑓′(𝑥)=limℎ→0𝑓(𝑥+ℎ)−𝑓(𝑥)ℎ We are considering 𝑓(𝑥)=𝑎𝑥, so 𝑓(𝑎+ℎ)=𝑥𝑥+ℎ. Making these substitutions in the derivative definition then gives us 𝑓′(𝑥)=𝑑𝑑𝑥𝑎𝑥=limℎ→0𝑎𝑥+ℎ−𝑎𝑥ℎ=limℎ→0𝑎𝑥𝑎ℎ−𝑎𝑥ℎ=limℎ→0𝑎𝑥(𝑎ℎ−1)ℎ[𝑎𝑥isunaffectedbythelimit]=𝑎𝑥limℎ→0𝑎ℎ−1ℎ
The preceding equation, combined with the discussion in Exploration 1, provides one way to define the number e:
Definition of e
e is defined to be the number such that
limℎ→0𝑒ℎ−1ℎ=1
To five digits, this number is 𝑒=2.71828.
The calculator below plots 𝑎ℎ−1ℎ versus h,
so you can see visually what the values are as limℎ→0.
Use the slider beneath the calculator to change the value of a: you'll find that for 𝑎=𝑒≈2.72, the limit as ℎ→0 equals 1.
With this definition of e in place, we have the key result we discovered in Exploration 1:
Derivative of 𝑒𝑥
𝑑𝑑𝑥𝑒𝑥=𝑒𝑥
We all love this particular derivative, since it's so easy to remember!
One reason e appears so often in describing physical phenomena
More importantly, this result is the first indication of why the number e appears so often when we describe physical phenomena: The rate of change of the function 𝑓(𝑥)=𝐶𝑒𝑥 is proportional to the value of the function itself. For instance, the way that bacterial growth occurs is that each bacterial cell subdivides so 1 becomes 2, and 2 become 4, and so on. That means that the rate at which they multiply at a given moment, 𝑑𝑁𝑑𝑡, is directly proportional to how many are present at that time, 𝑁(𝑡): they grow at twice the rate when there are 2 present as when there are 1, and four times the rate when there are 4 present instead of 1, and sixteen times the rate when there are 16 present than when there is 1. That is, 𝑑𝑁𝑑𝑡=𝑘𝑁(𝑡), where the constant k depends on the particular type of bacteria and how quickly they subdivide. This direct proportionality between 𝑑𝑁𝑑𝑡
and 𝑁(𝑡) leads straight to the bacterial growth equation 𝑁(𝑡)=𝑁0𝑒𝑘𝑡. We'll explore this and related ideas in much more depth later. For now, you might choose to marvel at how this number e has the remarkable property that 𝑑𝑑𝑥𝐶𝑒𝑥=𝐶𝑒𝑥.
Derivative of 𝑎𝑥 for other values of 𝑎
For completeness, here is a result that we'll be able to prove easily in a few screens. For now, we state the result and provide a more cumbersome proof in the Show/Hide box immediately below.
Derivative of 𝑎𝑥
For any value of 𝑎>0:𝑑𝑑𝑥𝑎𝑥=𝑎𝑥⋅ln𝑎
We'll be able to develop this result quite easily once we have the "Chain Rule" tool in our repertoire
a few screens from now. In the meantime:
The first few lines duplicate what we did above, since we once again start with the definition of the derivative applied to the function 𝑓(𝑥)=𝑎𝑥.
This result matches what we saw in Exploration 1 above: the function 𝑓(𝑥)=𝑎𝑥 and its derivative have the same-shaped curve because they differ only by the (constant) factor ln𝑎. And since ln𝑒=1, when 𝑎=𝑒 the function 𝑓(𝑥)=𝑒𝑥 and its derivative are identical.
The Desmos calculator below let's you examine the function 𝑓(𝑥)=𝑎𝑥 (solid curve) and its derivative 𝑓′(𝑥)=𝑎𝑥⋅ln𝑎 (dashed curve) on the same plot, as we did in Exploration 1. Now there is a slider beneath the graph that lets you see what happens for various values of a.
Practice Problems for the derivative of exponential functions
Practice Problem 1
Consider the function 𝑓(𝑥)=8𝑥. Then 𝑓′(𝑥)=
(A)8𝑥(B)7𝑥(C)8𝑥⋅ln8(D)8𝑥−1(E)noneofthese
We found above that when 𝑓(𝑥)=𝑎𝑥,𝑓′(𝑥)=𝑎𝑥⋅ln𝑎.
Hence given 𝑓(𝑥)=8𝑥,𝑓′(𝑥)=8𝑥⋅ln8⟹(C)✓
Practice Problem 2
Consider the function 𝑓(𝑥)=𝑒𝑥. Then 𝑑𝑓𝑑𝑥=
(A)𝑒(B)𝑒𝑥(C)𝑒𝑥−1(D)(𝑒−1)𝑥(E)noneofthese
This is one you will quickly remember:
𝑑𝑑𝑥𝑒𝑥=𝑒𝑥⟹(B)✓
Practice Problem 3
An equation for the tangent line to the curve 𝑦=𝑒𝑥 at 𝑥=2 is
Find the 𝑦-value of the point of interest: 𝑦0=𝑓(𝑎).
Find the value of the derivative at the point of interest, 𝑓′(𝑎).
Use these two pieces of information, 𝑚tangent=𝑓′(𝑎), and the point of interest (𝑥0,𝑦0), to write the equation of the tangent line in Point-Slope form:
Tangent Line to a Curve:𝑦−𝑦0=𝑚tangent(𝑥−𝑥0)𝑦−𝑓(𝑎)=𝑓′(𝑎)(𝑥−𝑎)
Step 1. Find the 𝑦-value of the point of interest: 𝑦0=𝑓(𝑎).
We're interested in the point 𝑥=2, so have 𝑥0=𝑎=2. Since 𝑦=𝑓(𝑥)=𝑒𝑥,
𝑦0=𝑓(2)=𝑒2
Hence we're looking for the tangent line to the curve at the point (2,𝑒2).◂
Step 2. Find the value of the derivative at the point of interest, 𝑓′(𝑎).
Since 𝑓(𝑥)=𝑒𝑥,
𝑓′(𝑥)=𝑒𝑥
Hence
𝑓′(2)=𝑒2◂
Step 3. Use these two pieces of information to write the equation of the tangent line. (See the Show/Hide box 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−𝑒2=𝑒2(𝑥−2)⟹(A)✓
Practice Problem 4
Consider the function 𝑓(𝑥)=3𝑥. You know that 𝑓(2)=32=9.
Use a linear approximation to estimate the value of 32.01.
Using a calculator we find that our approximation is, to four decimal places, 32.01≈9+0.09ln3=9.0989.
For comparison purposes, the actual value, to four decimal places, is 32.01=9.0994.
What questions or thoughts do you have about the material on this screen, or any other Calculus-related item? How are these problems: easy? challenging? Please let the Community know on our Forum!
The Upshot
The easiest derivative of all to remember: 𝑑𝑑𝑥𝑒𝑥=𝑒𝑥.
Almost as easy: 𝑑𝑑𝑥𝑎𝑥=𝑎𝑥⋅ln𝑎
On the
next screen
, we'll add the derivatives of two trig functions to our repertoire: 𝑑𝑑𝑥sin𝑥 and 𝑑𝑑𝑥cos𝑥.
