A.2 Derivative of Exponential Functions

Calculus Home Calculating Derivatives A. Derivatives of Polynomials, Exponential, and Trig Functions A.2 Derivative of Exponential Functions

On this screen we’re going to use Desmos to examine the derivative of exponential functions like $2^x,$ $3.25^x,$ and such. The following Exploration examines one very special such exponential function.

EXPLORATION 1: Derivative of $a^x$ and discovery of . . .

Let’s see how the discovery we made in Exploration 1 follows from the definition of the derivative as applied to the exponential function $f(x) = a^x.$ Recall the definition of the derivative from the preceding Chapter, “The Derivative”:
\[\textbf{Definition of the Derivative:}\qquad f'(x) = \lim_{h \to 0}\frac{f(x+h) – f(x)}{h} \] Since we have $f(x) = a^x,$ $f(a+h) = x^{x+h}.$ Making these substitutions into the definition of the derivative then gives us
\begin{align*}
f'(x) = \dfrac{d}{dx}a^x &= \lim_{h \to 0}\frac{a^{x+h} – a^x}{h} \\[8px] &= \lim_{h \to 0}\frac{a^x a^h – a^x}{h} \\[8px] &= \lim_{h \to 0}\frac{a^x \left(a^h – 1 \right)}{h} \quad \left[a^x\text{ is unaffected by the limit} \right] \\[8px] &= a^x \lim_{h \to 0}\frac{a^h – 1}{h}
\end{align*}
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
\[\lim_{h \to 0}\frac{e^h – 1}{h} = 1\] To five digits, this number is $e = 2.71828.$

Show/Hide Desmos calculator to explore limit for different values of a

The calculator below plots $\dfrac{a^h – 1}{h}$ versus h, so you can see visually what the values are as $\displaystyle{\lim_{h \to 0}}.$ Use the slider beneath the calculator to change the value of a: you’ll find that for $a = e \approx 2.72,$ the limit equals 1.

Graph of $\displaystyle{\lim_{h \to 0}\frac{a^h – 1}{h}}$ versus h

Use the slider to change the value of a:

Currently $a = 1.95.$

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With this definition of e in place, we have the key result we discovered in Exploration 1:

Derivative of $e^x$
\[\dfrac{d}{dx}e^x = e^x\]

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 very often when we describe physical phenomena: the rate at which the function $f(x) = Ce^x$ changes for a given value of x is proportional to the value of the function itself. Bacterial growth is one such situation: since each bacterial cell subdivides so 1 becomes 2, and 2 become 4, and so on, the rate at which they multiply at a given moment, $\dfrac{dN}{dt},$ depends on how many are present at that time. $N(t).$ That is, $\dfrac{dN}{dt} = kN(t),$ where the constant k depends on the particular type of bacteria and how quickly they subdivide. This direct proportionality between $\dfrac{dN}{dt}$ and $N(t)$ leads straight to the bacterial growth equation $N(t) = N_0 e^{kt}.$ 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 $\dfrac{d}{dx}Ce^x = Ce^x.$

Derivative of $a^x$ 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 $a^x$
For any value of $a \gt 0:$
\[\dfrac{d}{dx}a^x = a^x \cdot \ln a\]

Show/Hide development of the preceding result

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 $f(x) = a^x.$
\begin{align*}
f'(x) &= \lim_{h \to 0}\frac{a^{x+h} – a^x}{h} \\[8px] &= \lim_{h \to 0}\frac{a^x a^h – a^x}{h} \\[8px] &= \lim_{h \to 0}\frac{a^x \left(a^h – 1 \right)}{h} \quad \left[a^x\text{ is unaffected by the limit} \right] \\[8px] &= a^x \, \lim_{h \to 0}\frac{a^h – 1}{h}
\end{align*}
To proceed, recall that $a = e^{\ln a}.$
\begin{align*}
&= a^x \, \lim_{h \to 0}\frac{\left(e^{\ln a} \right)^h – 1}{h}\phantom{\quad \left[a^x\text{ is unaffected by the limit} \right]} \\[8px] &= a^x \, \lim_{h \to 0}\frac{e^{h\ln a} – 1}{h} \\[8px] \end{align*}
To proceed further, we make the substitution $u = h\ln a \implies h = \dfrac{u}{\ln a}.$
Note that $h \to 0 \implies u \to 0.$
Let’s make the substitution $h = \dfrac{u}{\ln a}:$
\begin{align*}
&= a^x \, \lim_{u \to 0}\frac{e^{u}\, – 1}{\frac{u}{\ln a}} \phantom{\quad \left[a^x\text{ is unaffected by the limit} \right]} \\[8px] &= a^x \, \overbrace{\lim_{u \to 0}\frac{e^{u}\, – 1}{u}}^{1} \, \ln a \quad \left[ \text{Recall from the definition of }e: \lim_{u \to 0}\frac{e^u\, – 1}{u} = 1 \right] \\[8px] f'(x) = \dfrac{d}{dx}a^x &= a^x \, \ln a \quad \cmark
\end{align*}

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This result matches what we saw in Exploration 1: the function $f(x) = a^x$ and its derivative have the same-shaped curve because they differ only by the (constant) factor $\ln a.$ And since $\ln e = 1,$ when $a = e$ the function $f(x) = e^x$ and its derivative are identical.

The Desmos calculator below let’s you examine the function $a^x$ and its derivative 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.

Graph of $f(x) = a^x$ and $f'(x) = a^x \cdot \ln a$ versus x

Use the slider to change the value of a:

Currently $f(x) = 1.95^x$

Practice Problems

Practice Problem #1

Consider the function $f(x) = 8^x.$ Then $f'(x) = $

\begin{array}{lllll} \text{(A) }8^x && \text{(B) }7^x && \text{(C) }8^x \cdot\ln 8 && \text{(D) }8^{x-1} && \text{(E) none of these } \end{array}

Show/Hide Solution

We found above that when $f(x) = a^x,$ $f'(x) = a^x \cdot \ln a$.
Hence given $f(x) = 8^x,$ \[f'(x) = 8^x \cdot \ln 8 \implies \text{ (C)} \quad \cmark \]

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Practice Problem #2

Consider the function $f(x) = e^x.$ Then $\dfrac{df}{dx} = $

\begin{array}{lllll} \text{(A) }e && \text{(B) }e^x && \text{(C) }e^{x-1} && \text{(D) }(e-1)^x && \text{(E) none of these} \end{array}

Show/Hide Solution

This is one you will quickly remember: \[\dfrac{d}{dx}e^x = e^x \implies \text{ (B)} \quad \cmark \]

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Practice Problem #3

An equation for the tangent line to the curve $y = e^x$ at $x = 2$ is

\begin{array}{lll} \text{(A) }y\, -\, e^2 = e^2(x-2) && \text{(B) }y\, -\, e = e(x-2) && \text{(C) }y\, -\, 2 = e^2(x-e^2) \end{array}

\begin{array}{ll} \text{(D) }y + e^2 = e^2(x+2) && \text{(E) none of these} \end{array}

Show/Hide Solution

Show/Hide steps to find equation for the tangent line to a curve

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

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:

Tangent Line to a Curve:

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

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Step 1. Find the $y$-value of the point of interest: $y_0 = f(a).$
We’re interested in the point $x=2,$ so have $x_0 = a = 2.$ Since $y = f(x) = e^x,$ \[y_0 = f(2) = e^2 \] Hence we’re looking for the tangent line to the curve at the point $(2, e^2). \,\blacktriangleleft$ Find the value of the derivative at the point of interest, $f'(a).$
Since $f(x) = e^x,$ \[f'(x) = e^x \] Hence \[f'(2) = e^2 \, \blacktriangleleft\] 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 \[y\, -\, e^2 = e^2(x-2) \implies \text{( A)} \quad \cmark \]

Graph showing the function f(x) = e^x, and the tangent line to its curve at x=2: y - e^2 = e^2(x - 2).

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Practice Problem #4

Consider the function $f(x) = 3^x.$ You know that $f(2) = 3^2 = 9.$

Use a linear approximation to estimate the value of $3^{2.01}.$

[Reminder: We developed linear approximations here.]

\begin{array}{lllll} \text{(A) }9 + 3 \ln 3 && \text{(B) }2 + 0.09 \ln 3 && \text{(C) }0.09 \ln 3 && \text{(D) }9 + 0.09 \ln 3 && \text{(E) none of these} \end{array}

Show/Hide Solution

ur solution requires the rate at which the function $f(x) = 3^x$ changes at $x=2.$ That rate is, of course, the derivative’s value at $x=2$: \begin{align*} f(x) &= 3^x \\[8px] f'(x) &= 3^x \cdot \ln 3 \\[8px] f'(2) &= 3^2 \cdot \ln 3 = 9 \ln 3 \; \blacktriangleleft \end{align*} We can then use that rate \begin{align*} \overbrace{f(2+dx)}^{\text{value at }2+dx} &\approx \overbrace{f(2)}^{f \text{ value at }2} + \overbrace{\text{small change }df}^{(\text{rate at }x=2)\cdot(dx)} \\[8px] f(2+0.01) &\approx 9 + (9 \ln 3) \cdot (0.01) \\[8px] &\approx 9 + 0.09 \ln 3 \implies \text{ (D)} \quad \cmark \end{align*} Using a calculator we find that our approximation is, to four decimal places, $3^{2.01} \approx 9 + 0.09 \ln 3 = 9.0989.$
For comparison purposes, the actual value, to four decimal places, is $3^{2.01} = 9.0994.$

You can use the Desmos calculator below to visualize this calculation for various values of dx. As with the similar calculators in Chapter 1, you must first zoom in until the small Leibniz triangle appears near the point $(2, 9).$

Graph of $f(x) = 3^x$ versus x

Use the slider to change the value of dx:

Currently dx = 0.01

Check/uncheck box to hide/show the $y=3^x$ curve.

The linear approximation for the current value of dx is:

\begin{align*} \overbrace{f(2+dx)}^\text{$f$ value at $2+dx$} &\approx \overbrace{f(2)}^{\text{$f$ value at }2} + \overbrace{\text{ small change }df}^{\text{(rate at $x=2$)} * \, (dx)} \\[8px] f(2+0.01) &\approx 9+ 9 (\ln 3)(0.01) \\[8px] \end{align*}

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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: $\dfrac{d}{dx}e^x = e^x.$
Almost as easy: $\dfrac{d}{dx}a^x = a^x \cdot \ln a$

On the next screen, we’ll add the derivatives of two trig functions to our repertoire: $\dfrac{d}{dx}\sin x$ and $\dfrac{d}{dx}\cos x.$

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