The derivative of exponential functions

Recall: Prior Knowledge

The exponential function has the general form $f(x)=b^x$ , where $b$ is the base. An often used base is $e=2.71828182845...$ (Euler's number), that is, $f(x)=e^x$ . It turns out that the derivative of this exponential function is particularly simple.

Also recall that $\log_b(b)=1$ for all bases $b$ , and in particular that $\ln(e)=1$ , as $\ln$ is a shorthand for $\log_e$ .

Theorem 1

The derivative of an exponential function is the same exponential function, but multiplied by some constant:

Equation 1

f(x)=b^x \overset{\prime}{\rightarrow} f'(x)=\ln(b)\cdot b^x

Derivative of exponential function

In particular, it follows from this rule that

Equation 2

f(x)=e^x \overset{\prime}{\rightarrow} f'(x)= e^x

Proof

We just give an argument why this is plausible, using graphical differentiation. We use a concrete example: $f(x)=e^x$ . Sketching the graph of $f(x)=e^x$ and then determining graphically the graph of the derivative $f'$ , we see that the graph of $f'$ looks very similar to the graph of $f$ ; that is, we suspect that $f'(x)=e^x$ as well. A more formal, mathematical proof is shown in the collapsed exercise below, but it is not expected to know it.

The reason for the appearance of the factor $\ln(b)$ will be explained later.

Figure 1

The graphical derivative of $f(x)=e^x$ suggests that the derivative of $e^x$ is also $e^x$ .

Warning

A common mistake is to apply the power rule for the exponential function; for example, for $f(x)=2^x$ , you might incorrectly write $f'(x)=x\cdot 2^{x-1}$ . Always check if the function you want to derive is an exponential function ( $2^x$ ) or a power function ( $x^2$ ).

Example 1

Calculate the derivative using the exponential rule we just learned about:

$f(x)=2^x \rightarrow f'(x)=\ln(2)\cdot 2^x = 0.69314... \cdot 2^x$
$f(x)=e^x \rightarrow f'(x)=\underbrace{\ln(e)}_{=1} \cdot e^x = e^x$

So we can use the general rule as well to see that the derivative of $e^x$ is again $e^x$ .

Here are some more formal arguments for proving the exponential rule, but you can skip them and read them later.

Show

Exercise 1

Use the differential quotient to show that the derivative of $f(x)=e^x$ is $f'(x)=e^x$ .

Hint:

\frac{e^h-1}{h} \rightarrow 1\quad (\text{for } h\rightarrow 0)

Solution

The differential quotient is

f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}

We have

\begin{aligned} \frac{f(x+h)-f(x)}{h} & = \frac{e^{x+h}-e^x}{h} \\ & = \frac{e^x\cdot e^h-e^x}{h} \\ & = \frac{e^x\cdot (e^h-1)}{h} \\ & = \frac{e^h-1}{h}\cdot e^x \end{aligned}

Now we have a problem because if we insert $h=0$ , we get $\frac{e^h-1}{h}=\frac{0}{0}$ , and there is no obvious way to get rid of the $h$ in the denominator. So let us find out numerically what is going on. Inserting smaller and smaller values for $h$ , we see that the quotient $\frac{e^h-1}{h}$ converges towards $1$ .

Thus we see that

\frac{e^h-1}{h} \rightarrow 1\quad (\text{for } h\rightarrow 0)

and therefore it is indeed $f'(x)=e^x$ .

Exercise 2

Use a similar proof to show that

f(x)=2^x \rightarrow f'(x)=\ln(2)\cdot 2^x

Hint: $\ln(2)=0.6931471805599453...$

Solution

The difference quotient is

\begin{aligned} \frac{f(x+h)-f(x)}{h} & = \frac{2^{x+h}-2^x}{h} \\ & = \frac{2^x\cdot 2^h-2^x}{h} \\ & = \frac{2^x\cdot (2^h-1)}{h} \\ & = \frac{2^h-1}{h}\cdot 2^x \end{aligned}

Inserting smaller and smaller values for $h$ , we see that the quotient $\frac{2^h-1}{h}$ converges towards $\ln(2)$ .

Thus we have $f'(x)=\ln(2)\cdot 2^x$ .

Exercise 3

Determine the derivative of $f$ :
1. $f(x)=7^x$
2. $f(x)=10 e^x$
3. $f(x)=2 e^x-4\cdot 6^x$
4. $f(x)=x^4-4^x$
Find the equation of the tangent with slope $0.5$ to the graph of $f(x)=3\cdot 5^x$ .
Consider the function $f(x)=3e^x-4$ :
1. Where does the graph of $f$ intersect the $x$ -axis, and where the $y$ -axis?
2. What is the angle of intersection between the graph of $f$ and the $x$ -axis ( $x$ -intercept)?
3. What is the angle of intersection between the graph of $f$ and the $y$ -axis ( $y$ -intercept)?

Solution

1. $f'(x)=\ln(7)\cdot 7^x$
2. $f'(x)=10e^x$
3. $f'(x)=2 e^x - 4\ln(6)\cdot 6^x$
4. $f'(x)=4x^3-\ln(4)\cdot 4^x$
First, we need to find the point $A(x\vert y)$ where the tangent touches the graph. Thus, find $x$ with
$f'(x)=3\ln(5)\cdot 5^x = 0.5$
and thus
$5^x=\frac{0.5}{3\ln(5)}$
Taking the logarithm base 5 on both sides, we get
$x = \log_5\left(\frac{0.5}{3\ln(5)}\right) \approx -1.409$
Thus, $y=f(-1.409)=3\cdot 5^{-1.409} \approx 0.311$ and $A(-1.409\vert 0.311)$ .

Second, to find the equation $t(x)=ax+b$ of the tangent, we note that $a=0.5$ , and therefore
$t(-1.409)=0.5\cdot (-1.409)+b = 0.311$
and it follows $b \approx 1.02$ . Thus, we have $t(x)=\underline{0.5x+1.02}$ .
$f'(x)=3e^x$

(a)+(c): $y$ -intercept: $f(0)=3e^0-4=3-4=\underline{-1}$ . The slope of the tangent at this point is $a=f'(0)=3e^{0}=3$ , which corresponds to an angle of $\alpha=\arctan(a)=\arctan(3) \approx 71.565^\circ$ with respect to the $x$ -axis. With respect to the $y$ -axis, the angle is $\beta=90^\circ - \alpha=\underline{18.435^\circ}$ .

(a)+(b): $x$ -intercept: Find $x$ with $f(x)=3e^x-4=0 \rightarrow e^x=\frac{4}{3} \rightarrow x=\ln(4/3) \approx \underline{0.288}$ . The slope of the tangent there is $a=f'(\ln(4/3))=3e^{\ln(4/3)}=4$ , which corresponds to an angle of $\alpha=\arctan(a)=\arctan(4) \approx \underline{75.964^\circ}$ relative to the $x$ -axis.