On this screen we’re going 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.

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:*

\[\lim_{h \to 0}\frac{e^h – 1}{h} = 1\] 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

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

Use the slider to change the value of *a*:

Currently $a = 1.95.$

Currently $a = 1.95.$

[collapse]

With this definition of

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

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 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.

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 $f(x) = 1.95^x$

Currently $f(x) = 1.95^x$

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}

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}

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}

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}

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 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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