Derivatives of Exponential Functions

You’ve already met the rule $\dfrac{d}{dx}\big[e^x\big] = e^x$ on the derivatives of sine, cosine, eˣ, and ln x page. This page shows where it comes from: every exponential function has a slope that is a fixed multiple of its own height, and $e$ is the one base where that multiple is exactly $1$. From there you’ll get the derivative of any exponential, $a^x$ or $A \cdot b^{kt}$, and use it to find how fast populations, investments, and radioactive samples are changing.

Key ideas

The slope of a^x is proportional to a^x

Take $f(x) = a^x$, with $a \gt 0$ and $a \ne 1$. Use the definition of the derivative and the exponent law $a^{x + h} = a^x \cdot a^h$:

$$f'(x) = \lim_{h \to 0} \frac{a^{x + h} - a^x}{h} = \lim_{h \to 0} \frac{a^x\big(a^h - 1\big)}{h} = a^x \cdot \lim_{h \to 0} \frac{a^h - 1}{h}$$

The limit at the end doesn’t contain $x$ at all. It’s just a number that depends on the base $a$. Call it $k$:

$$f'(x) = k \cdot a^x, \qquad \text{where } k = \lim_{h \to 0} \frac{a^h - 1}{h}$$

So the slope of an exponential graph at any point is a constant times its height. In other words, the ratio $\dfrac{f'(x)}{f(x)}$ is the same constant $k$ everywhere. Since $f'(0) = k \cdot a^0 = k$, the constant is also the slope of the graph where it crosses the $y$-axis.

Estimating k with a table

You can’t simplify $\dfrac{a^h - 1}{h}$ algebraically, but a calculator can estimate the limit. Use smaller and smaller values of $h$:

$h$	$0.1$	$0.01$	$0.001$	$0.0001$
$\dfrac{2^h - 1}{h}$	$0.7177$	$0.6956$	$0.6934$	$0.6932$
$\dfrac{3^h - 1}{h}$	$1.1612$	$1.1047$	$1.0992$	$1.0987$

(Values rounded to 4 decimal places.) So for $2^x$, $k \approx 0.693$, and for $3^x$, $k \approx 1.099$:

$$\frac{d}{dx}\big[2^x\big] \approx 0.693 \cdot 2^x, \qquad \frac{d}{dx}\big[3^x\big] \approx 1.099 \cdot 3^x$$

The base where k = 1 is e

For base $2$ the constant is less than $1$, and for base $3$ it’s more than $1$. Somewhere between $2$ and $3$ there’s a base whose constant is exactly $1$. That base is the number

$$e \approx 2.718\,28$$

If you graph $y = a^x$ with a slider for $a$ (Desmos or GeoGebra work well) and compare the slope of the tangent with the height of the graph, they match only when $a = e$. For $f(x) = e^x$:

$$f'(x) = e^x$$

In words: the slope of the tangent at any point equals the height of the graph at that point. At $(1, e)$ the slope is $e$; at $(2, e^2)$ the slope is $e^2 \approx 7.389$.

At $(0, 1)$ the tangent slopes are about $0.693$ for $y = 2^x$, exactly $1$ for $y = e^x$, and about $1.099$ for $y = 3^x$.

The constant is ln a

Compare the constants with natural logarithms: $\ln 2 \approx 0.6931$ and $\ln 3 \approx 1.0986$. They match the table. This works for every base:

$a$	$1.7$	$2$	$2.5$	$e$	$3$	$3.5$
$k = \ln a$	$0.531$	$0.693$	$0.916$	$1$	$1.099$	$1.253$

Here’s why. Since $a = e^{\ln a}$, you can write $a^x = e^{(\ln a)x}$. The chain rule (inside function $(\ln a)x$, whose derivative is the constant $\ln a$) gives

$$\frac{d}{dx}\big[a^x\big] = e^{(\ln a)x} \cdot \ln a = a^x \ln a$$

This includes $e^x$ as a special case, because $\ln e = 1$.

What the graph of the derivative looks like

Since $f'(x) = (\ln a) \cdot f(x)$, the graph of $f'$ is the graph of $f$ stretched vertically by a factor of $\ln a$. So $f'$ is also an exponential function with the same base:

• If $a \gt e$, then $\ln a \gt 1$: $f'$ sits above $f$ (an expansion).
• If $1 \lt a \lt e$, then $0 \lt \ln a \lt 1$: $f'$ sits below $f$ (a compression), as for $2^x$ in the figure.
• If $0 \lt a \lt 1$, then $\ln a \lt 0$: $f'$ is a reflection in the $x$-axis as well. That makes sense, because a decaying exponential is always decreasing.

The derivative of $y = 2^x$ is $y = (\ln 2) \cdot 2^x$: the same curve, compressed vertically by a factor of $\ln 2 \approx 0.693$.

Exponentials with the chain rule

Combine the rule with the chain rule whenever the exponent is more than just $x$:

Function	Derivative
$e^{kx}$	$k e^{kx}$
$e^{g(x)}$	$e^{g(x)} \cdot g'(x)$
$a^{g(x)}$	$a^{g(x)} \ln a \cdot g'(x)$
$A \cdot b^{kt}$	$A \cdot b^{kt} \cdot k \ln b$

The last line is the general growth or decay model. Notice that its derivative is $(k \ln b)$ times the original function: the rate of change is proportional to the amount present. That’s the defining feature of exponential growth and decay. A population of $10\,000$ grows twice as fast (in people per year) as the same kind of population of $5000$.

Worked examples

Example 1: Estimating a derivative numerically

Let $f(x) = 3^x$. Use the table of values of $\dfrac{3^h - 1}{h}$ above to estimate $f'(2)$. Then check your estimate with the rule $f'(x) = a^x \ln a$.

Solution. From the definition,

$$f'(2) = \lim_{h \to 0} \frac{3^{2 + h} - 3^2}{h} = 3^2 \cdot \lim_{h \to 0} \frac{3^h - 1}{h} \approx 9(1.0987) \approx 9.888$$

Check with the rule: $f'(x) = 3^x \ln 3$, so

$$f'(2) = 9\ln 3 \approx 9(1.098\,61) \approx 9.888$$

The two answers agree (to 3 decimal places). The exact value is $9\ln 3$.

Example 2: Differentiating exponential functions

Find the derivative of each function.

• (a) $y = 5^x$
• (b) $y = e^{-4x}$
• (c) $y = 3e^{x^2}$
• (d) $y = 2^{3x + 1}$

Solution.

(a) Use $\dfrac{d}{dx}\big[a^x\big] = a^x \ln a$: $\quad \dfrac{dy}{dx} = 5^x \ln 5$.

(b) The inside function is $-4x$, with derivative $-4$: $\quad \dfrac{dy}{dx} = -4e^{-4x}$.

(c) The inside function is $x^2$, with derivative $2x$:

$$\frac{dy}{dx} = 3e^{x^2} \cdot 2x = 6xe^{x^2}$$

(d) The inside function is $3x + 1$, with derivative $3$:

$$\frac{dy}{dx} = 2^{3x + 1} \ln 2 \cdot 3 = 3\ln 2 \cdot 2^{3x + 1}$$

Example 3: A tangent line

Find the equation of the tangent line to $y = 2^x$ at $x = 3$. Where does the tangent line cross the $x$-axis?

Solution. The point is $(3, 2^3) = (3, 8)$. The slope is

$$\frac{dy}{dx} = 2^x \ln 2 \quad\Rightarrow\quad m = 2^3 \ln 2 = 8\ln 2 \approx 5.545$$

The tangent line is

$$y - 8 = 8\ln 2\,(x - 3)$$

It crosses the $x$-axis where $y = 0$:

$$-8 = 8\ln 2\,(x - 3) \quad\Rightarrow\quad x - 3 = -\frac{1}{\ln 2} \quad\Rightarrow\quad x = 3 - \frac{1}{\ln 2} \approx 1.557$$

Example 4: Radioactive decay

A hospital patient receives $200$ mg of iodine-131, which has a half-life of $8$ days. The amount remaining after $t$ days is

$$A(t) = 200\left(\frac{1}{2}\right)^{t/8}$$

• (a) Find the rate of decay at $t = 0$ and at $t = 8$ days.
• (b) Show that $A'(t)$ is proportional to $A(t)$.
• (c) When is the amount decreasing at a rate of $2$ mg per day?

Solution.

(a) Use $a^{g(t)}$ with $a = \frac{1}{2}$ and $g(t) = \frac{t}{8}$. Remember $\ln \frac{1}{2} = -\ln 2$:

$$A'(t) = 200\left(\frac{1}{2}\right)^{t/8} \cdot \ln\frac{1}{2} \cdot \frac{1}{8} = -25\ln 2\left(\frac{1}{2}\right)^{t/8}$$ $$A'(0) = -25\ln 2 \approx -17.329, \qquad A'(8) = -25\ln 2 \cdot \frac{1}{2} = -12.5\ln 2 \approx -8.664$$

At the start, the iodine is decaying at about $17.329$ mg per day. After one half-life, half as much is left, and it’s decaying half as fast: about $8.664$ mg per day. (The negative sign means the amount is decreasing.)

(b) From part (a),

$$A'(t) = \frac{\ln \frac{1}{2}}{8} \cdot 200\left(\frac{1}{2}\right)^{t/8} = -\frac{\ln 2}{8} A(t) \approx -0.0866\,A(t)$$

So at any moment, the amount decreases at about $8.66\%$ of the current amount per day.

(c) Set $A'(t) = -2$. Using part (b), $-\dfrac{\ln 2}{8} A(t) = -2$, so $A(t) = \dfrac{16}{\ln 2} \approx 23.083$ mg. Now solve for $t$:

$$\begin{aligned} 200\left(\frac{1}{2}\right)^{t/8} &= \frac{16}{\ln 2} \\ \left(\frac{1}{2}\right)^{t/8} &= \frac{2}{25\ln 2} \\ \frac{t}{8}\ln\frac{1}{2} &= \ln\left(\frac{2}{25\ln 2}\right) && \text{take ln of both sides} \\ t &= \frac{8\ln\left(\frac{2}{25\ln 2}\right)}{\ln \frac{1}{2}} \approx 24.921 \end{aligned}$$

After about $24.921$ days (just under $25$ days), the iodine is decaying at $2$ mg per day.

Common mistakes

Using the power rule on an exponential. $\dfrac{d}{dx}\big[2^x\big]$ is not $x \cdot 2^{x - 1}$. The power rule is for a variable base and a constant exponent ($x^2$). Here the base is constant and the exponent is the variable, so you need $a^x \ln a$.

Forgetting the ln a. $\dfrac{d}{dx}\big[5^x\big] = 5^x \ln 5$, not $5^x$. Only base $e$ has no extra factor, because $\ln e = 1$. A quick check: the slope of $5^x$ at $x = 0$ should be about $1.609$, not $1$.

Forgetting the chain rule factor. $\dfrac{d}{dx}\big[e^{-4x}\big] = -4e^{-4x}$, not $e^{-4x}$. Whenever the exponent is anything other than plain $x$, multiply by the derivative of the exponent.

Changing the exponent. The derivative of $e^{x^2}$ is $2xe^{x^2}$. The exponent stays $x^2$; it does not become $2x$ or drop by one. The $2x$ comes out in front as a factor.

Dropping the negative sign for decay. For $b \lt 1$, $\ln b$ is negative, so the derivative is negative. If your “rate of decay” comes out positive, check $\ln \frac{1}{2} = -\ln 2$. In words, you can then say “decreasing at $17.329$ mg per day”.

Rounding too early. Keep $\ln 2$ or the full calculator value until the last step. Rounding $\ln 2$ to $0.7$ in Example 4 would give $A'(0) \approx -17.5$, which is wrong to 3 decimal places.

Practice

1. (Warm-up) Differentiate each function.

• (a) $y = 7^x$
• (b) $y = e^{5x}$
• (c) $y = 4e^{-0.2x}$

Solution

(a) $\dfrac{dy}{dx} = 7^x \ln 7$

(b) $\dfrac{dy}{dx} = 5e^{5x}$

(c) $\dfrac{dy}{dx} = 4e^{-0.2x} \cdot (-0.2) = -0.8e^{-0.2x}$

2. (Warm-up) Let $f(x) = 5^x$.

• (a) Evaluate $\dfrac{5^h - 1}{h}$ for $h = 0.01$, $h = 0.001$ and $h = 0.0001$, to 4 decimal places.
• (b) Use your values to estimate the constant $k$ in $f'(x) = k \cdot 5^x$, and compare it with $\ln 5$.

Solution

(a) $\dfrac{5^{0.01} - 1}{0.01} \approx 1.6225$, $\quad \dfrac{5^{0.001} - 1}{0.001} \approx 1.6107$, $\quad \dfrac{5^{0.0001} - 1}{0.0001} \approx 1.6096$.

(b) The values are settling down near $1.609$, so $k \approx 1.609$ and $f'(x) \approx 1.609 \cdot 5^x$. This matches $\ln 5 \approx 1.6094$.

3. (Warm-up) Explain, without a calculator, why the slope of the tangent to $y = e^x$ at the point $(2, e^2)$ is $e^2$. Then give that slope to 3 decimal places.

Solution

For $f(x) = e^x$, $f'(x) = e^x$: the slope at any point equals the height of the graph there. At $x = 2$ the height is $e^2$, so the slope is $e^2 \approx 7.389$.

4. (Core) Differentiate each function.

• (a) $y = 3^{2x - 1}$
• (b) $y = 10^{x^2}$
• (c) $y = x^2 e^{-3x}$

Solution

(a) The exponent $2x - 1$ has derivative $2$:

$$\frac{dy}{dx} = 3^{2x - 1} \ln 3 \cdot 2 = 2\ln 3 \cdot 3^{2x - 1}$$

(b) The exponent $x^2$ has derivative $2x$:

$$\frac{dy}{dx} = 10^{x^2} \ln 10 \cdot 2x = 2x\ln 10 \cdot 10^{x^2}$$

(c) Use the product rule, with the chain rule on $e^{-3x}$:

$$\frac{dy}{dx} = 2x \cdot e^{-3x} + x^2 \cdot \big(-3e^{-3x}\big) = xe^{-3x}(2 - 3x)$$

5. (Core) Find the equation of the tangent line to $y = 6 - 2e^{x - 1}$ at $x = 1$.

Solution

Point: $y = 6 - 2e^0 = 4$, so $(1, 4)$.

Slope: $\dfrac{dy}{dx} = -2e^{x - 1}$, so at $x = 1$ the slope is $-2e^0 = -2$.

$$y - 4 = -2(x - 1) \quad\Rightarrow\quad y = -2x + 6$$

6. (Core) The population of a town is modelled by $P(t) = 12\,000(1.025)^t$, where $t$ is the number of years after 2026.

• (a) Find $P'(t)$.
• (b) How fast is the population growing in 2026 and in 2036? Round to the nearest person per year.
• (c) Why is the growth rate larger in 2036?

Solution

(a) $P'(t) = 12\,000(1.025)^t \ln 1.025$

(b) In 2026, $t = 0$: $P'(0) = 12\,000\ln 1.025 \approx 296.3$, so about $296$ people per year.

In 2036, $t = 10$: $P'(10) = 12\,000(1.025)^{10}\ln 1.025 \approx 379.3$, so about $379$ people per year.

(c) The rate of growth is proportional to the population: $P'(t) = (\ln 1.025)P(t) \approx 0.0247\,P(t)$. By 2036 the population has grown to about $15\,361$, so it grows by more people each year.

7. (Core) You invest $2000 at $6\%$ per year, compounded annually, so the value after $t$ years is $A(t) = 2000(1.06)^t$ dollars.

• (a) Find the rate at which the investment is growing at $t = 0$, to the nearest cent per year.
• (b) When is the investment growing at $150 per year? Round to 2 decimal places.

Solution

(a) $A'(t) = 2000(1.06)^t \ln 1.06$, so $A'(0) = 2000\ln 1.06 \approx 116.54$. The investment is growing at about $116.54 per year at the start.

(b) Solve $A'(t) = 150$:

$$\begin{aligned} 2000\ln 1.06\,(1.06)^t &= 150 \\ (1.06)^t &= \frac{150}{2000\ln 1.06} \approx 1.287\,13 \\ t &= \frac{\ln(1.287\,13)}{\ln 1.06} \approx 4.33 \end{aligned}$$

After about $4.33$ years, the investment is growing at $150 per year.

8. (Challenge) Let $f(x) = e^{2x} - 6x$. Find the point where the graph has a horizontal tangent, and decide whether it’s a local maximum or minimum.

Solution

$$f'(x) = 2e^{2x} - 6 = 0 \quad\Rightarrow\quad e^{2x} = 3 \quad\Rightarrow\quad x = \frac{\ln 3}{2} \approx 0.549$$

The $y$-coordinate is $f\left(\frac{\ln 3}{2}\right) = e^{\ln 3} - 6 \cdot \frac{\ln 3}{2} = 3 - 3\ln 3 \approx -0.296$.

$f''(x) = 4e^{2x}$, and $f''\left(\frac{\ln 3}{2}\right) = 4(3) = 12 \gt 0$, so by the second derivative test the point $\left(\dfrac{\ln 3}{2},\ 3 - 3\ln 3\right) \approx (0.549, -0.296)$ is a local minimum.

9. (Challenge) For $f(x) = a^x$, the ratio $\dfrac{f'(x)}{f(x)}$ is constant.

• (a) For which base $a$ is $f'(x) = 2f(x)$ for every $x$?
• (b) For which base $a$ is $f'(x) = -f(x)$ for every $x$? Describe how the graphs of $f$ and $f'$ are related in this case.

Solution

(a) $\dfrac{f'(x)}{f(x)} = \ln a$, so we need $\ln a = 2$, which gives $a = e^2 \approx 7.389$. (Check: $e^{2x}$ has derivative $2e^{2x}$.)

(b) We need $\ln a = -1$, so $a = e^{-1} = \dfrac{1}{e} \approx 0.368$. Then $f(x) = e^{-x}$ and $f'(x) = -e^{-x}$: the graph of $f'$ is the reflection of the graph of $f$ in the $x$-axis.