Derivatives of Exponential Functions

Today’s class was a hunt. With a slider controlling $a$, your group graphed $f(x)=a^x$ next to its derivative and chased a question: for what base does the derivative land exactly on top of the function? At $a=2$ the derivative ran below; at $a=3$ it ran above. The prey was somewhere between — and the class cornered it near $2.718$. That number is called $e$, and it is the reason this page is short.

$\text{If }f(x)=e^x,\text{ then }{f}^{\prime }(x)=e^x$

A function that is its own derivative: at every point, the slope of the tangent equals the height of the graph. Let that be strange for a minute. Growth whose speed is its size is exactly what populations, investments, and decaying isotopes do — which is why $e$ shows up wherever things grow.

Every other base

The hunt revealed something about the bases that lost, too. For any $a>0$, $a\ne 1$, the derivative of $a^x$ is a vertical stretch of the original — the ratio $\frac{f^{\prime }(x)}{f(x)}$ is the same constant at every $x$. The constant has a name:

$\text{If }f(x)=a^x,\text{ then }{f}^{\prime }(x)=a^x\ln a$

Here $\ln x$ is the natural logarithm, ${\log }_{e}x$ — the inverse of $e^x$, exactly as ${\log }_{2}x$ inverts $2^x$. And the two facts are one fact: when $a=e$, the stretch factor is $\ln e=1$, and the derivative lands on the function. Every base is trying to be $e$; only one succeeds.

Worth checking with your own hands — Using Desmos shows you how, and an Estimation Duels instinct makes the numbers meaningful:

• Graph $f(x)=2^x$ and its derivative. Verify the ratio $\frac{f^{\prime }(x)}{f(x)}$ is near $0.693$ at several values of $x$ — then compute $\ln 2$.
• Evaluate $\frac{2^h-1}{h}$ for $h=0.1$, $0.01$, $0.001$. The march should be heading for that same $0.693$.
• Predict the stretch factor for $f(x)=10^x$ before you check it. Bigger or smaller than 1? Why must it be bigger?

With the chain rule along, composites like $e^{3x}$ fall too: derivative $3e^{3x}$, the inside reporting its rate as always. Exponential and Sinusoidal Derivatives Practice puts this to work on real growth and decay, and the other family of functions that models the world — the repeating kind — is next, in Derivatives of Sinusoidal Functions.

Curriculum connection

A2.5

determine, through investigation using technology, the graph of the derivative $f^{\prime }(x)$ or $\frac{dy}{dx}$ of a given exponential function [i.e., $f(x)=a^x$ $(a>0,a\ne 1)$] [e.g., by generating a table of values showing the instantaneous rate of change of the function for various values of $x$ and graphing the ordered pairs; by using dynamic geometry software to verify that when $f(x)=a^x$, $f^{\prime }(x)=kf(x)$], and make connections between the graphs of $f(x)$ and $f^{\prime }(x)$ or $y$ and $\frac{dy}{dx}$ [e.g., $f(x)$ and $f^{\prime }(x)$ are both exponential; the ratio $\frac{f^{\prime }(x)}{f(x)}$ is constant, or $f^{\prime }(x)=kf(x)$; $f^{\prime }(x)$ is a vertical stretch from the x-axis of $f(x)$] Sample problem: Graph, with technology, $f(x)=a^x$ $(a>0,a\ne 1)$ and $f^{\prime }(x)$ on the same set of axes for various values of $a$ (e.g., 1.7, 2.0, 2.3, 3.0, 3.5). For each value of $a$, investigate the ratio $\frac{f^{\prime }(x)}{f(x)}$ for various values of $x$, and explain how you can use this ratio to determine the slopes of tangents to $f(x)$.

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A2.6

determine, through investigation using technology, the exponential function $f(x)=a^x$ $(a>0,a\ne 1)$ for which $f^{\prime }(x)=f(x)$ (e.g., by using graphing technology to create a slider that varies the value of $a$ in order to determine the exponential function whose graph is the same as the graph of its derivative), identify the number $e$ to be the value of $a$ for which $f^{\prime }(x)=f(x)$ [i.e., given $f(x)=e^x$, $f^{\prime }(x)=e^x$], and recognize that for the exponential function $f(x)=e^x$ the slope of the tangent at any point on the function is equal to the value of the function at that point Sample problem: Use graphing technology to determine an approximate value of $e$ by graphing $f(x)=a^x$ $(a>0,a\ne 1)$ for various values of $a$, comparing the slope of the tangent at a point with the value of the function at that point, and identifying the value of $a$ for which they are equal.

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A2.7

recognize that the natural logarithmic function $f(x)={\log }_{e}x$, also written as $f(x)=\ln x$, is the inverse of the exponential function $f(x)=e^x$, and make connections between $f(x)=\ln x$ and $f(x)=e^x$ [e.g., $f(x)=\ln x$ reverses what $f(x)=e^x$ does; their graphs are reflections of each other in the line $y=x$; the composition of the two functions, $e^{\ln x}$ or $\ln e^x$, maps $x$ onto itself, that is, $e^{\ln x}=x$ and $\ln e^x=x$]

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A2.8

verify, using technology (e.g., calculator, graphing technology), that the derivative of the exponential function $f(x)=a^x$ is $f^{\prime }(x)=a^x\ln a$ for various values of $a$ [e.g., verifying numerically for $f(x)=2^x$ that $f^{\prime }(x)=2^x\ln 2$ by using a calculator to show that ${\lim }_{h\to 0}\frac{(2^h-1)}{h}$ is $\ln 2$ or by graphing $f(x)=2^x$, determining the value of the slope and the value of the function for specific x-values, and comparing the ratio $\frac{f^{\prime }(x)}{f(x)}$ with $\ln 2$] Sample problem: Given $f(x)=e^x$, verify numerically with technology using ${\lim }_{h\to 0}\frac{(e^{x+h}-e^x)}{h}$ that $f^{\prime }(x)=f(x)\ln e$.

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