Calculus 1

Derivatives of exponential functions

The rules so far differentiate powers of x, where the variable sits in the base. What happens when the variable moves up into the exponent, as in 2 to the x, and why does one base, e, come out as its own derivative?

You need

Change of baseAlgebra 2

Unlocks

Antiderivatives of trig, exponential and log functionsCalculus 1 The common Maclaurin seriesCalculus 2

Where you are 3 in Derivatives of trig, exponential and log functions

Calculus8

Derivatives13

Derivatives of trig, exponential and log functions3

Derivatives of exponential functions

Derivatives of the trigonometric functions Derivatives of logarithms and logarithmic differentiation

Why this exists

The power rule differentiates x^3, where the base varies and the exponent is fixed. It says nothing about 3^x, where the base is fixed and the exponent varies, and exponential functions are exactly what grows money, populations and bacteria, and what shrinks a radioactive sample. To know how fast such a quantity is changing at an instant, the derivative of a^x has to be found from the definition of the derivative itself.

That calculation turns up something no polynomial does: the derivative of an exponential is the same exponential multiplied by a constant. The constant depends on the base, and exactly one base makes it 1. That base is the number e, and the constant for every other base turns out to be its natural logarithm, \ln a.

Definition

Let a > 0. Write the difference quotient of f(x) = a^x and use the product of powers rule, a^{x + h} = a^x a^h:

\begin{aligned} \frac{a^{x + h} - a^x}{h} &= \frac{a^x a^h - a^x}{h} \\ &= a^x \cdot \frac{a^h - 1}{h}. \end{aligned}

The factor a^x does not involve h, so it passes outside the limit by the constant multiple law. What is left is a limit that depends on a alone. Name it

L(a) = \lim_{h \to 0} \frac{a^h - 1}{h}.

It is the derivative of a^x at x = 0, since a^0 = 1: the slope of y = a^x where it crosses the y-axis.

The number e is the base for which this slope is exactly 1:

L(e) = \lim_{h \to 0} \frac{e^h - 1}{h} = 1.

That this base is the same e \approx 2.71828 that continuous compounding produces, the number \displaystyle \left(1 + \frac{1}{n}\right)^n approaches, was argued in the natural exponential function; it is taken as given here. The natural logarithm \ln a is the logarithm to base e, so e^{\ln a} = a for every a > 0. By Change of base, \ln a = \dfrac{\log a}{\log e}, which is how it was computed before calculators had an \ln key.

Intuition

The factoring says the slope of y = a^x at any point is its height times its slope at x = 0. Where the curve is twice as high it is twice as steep. A quantity whose rate of change is proportional to its size is exactly what an exponential models: a canister holding twice as much radon loses it twice as fast.

So everything hangs on the one number L(a). Here is \dfrac{a^h - 1}{h} for a = 2 and a = 3, rounded to four decimal places:

h	\dfrac{2^h - 1}{h}	\dfrac{3^h - 1}{h}
0.1	0.7177	1.1612
0.01	0.6956	1.1047
0.001	0.6934	1.0992
-0.001	0.6929	1.0980
-0.01	0.6908	1.0926
-0.1	0.6697	1.0404

From both sides the values close in on L(2) \approx 0.693 and L(3) \approx 1.099. The curve y = 2^x crosses the y-axis a little less steeply than slope 1, and y = 3^x a little more steeply, so some base between 2 and 3 crosses at slope exactly 1. That base is e. A calculator also gives \ln 2 \approx 0.6931 and \ln 3 \approx 1.0986, matching the table; the next section proves this is no coincidence.

Figure 1The tangent to y = 2^x at (3, 8) has slope 8 \ln 2 \approx 5.545 (Example 3). It meets the x-axis \displaystyle \frac{1}{\ln 2} \approx 1.443 units to the left of x = 3.

Results

Proportional to itself. For every base a > 0,

\frac{d}{dx}\big[a^x\big] = a^x \cdot L(a).

The natural exponential is its own derivative.

\frac{d}{dx}\big[e^x\big] = e^x.

Every other base brings a factor of \ln a. For every a > 0, L(a) = \ln a, so

\frac{d}{dx}\big[a^x\big] = a^x \ln a.

For a > 1, \ln a > 0 and the function increases; for 0 < a < 1, \ln a < 0 and it decreases; for a = 1 the function is the constant 1 and the formula gives 0.

A linear exponent. For every constant k,

\frac{d}{dx}\big[e^{kx}\big] = k e^{kx}.

Why it is true

The factoring, proved. The identity in the Definition holds for every h \ne 0. If L(a) exists, the constant multiple law gives \dfrac{d}{dx}\big[a^x\big] = a^x L(a) at every x. With a = e and L(e) = 1 this is \dfrac{d}{dx}\big[e^x\big] = e^x.

Why L(a) = \ln a, proved without the chain rule. Let a > 0 and a \ne 1, so \ln a \ne 0. Since a = e^{\ln a}, the power of a power rule gives a^h = e^{h \ln a}. Put k = h \ln a. Then

\begin{aligned} \frac{a^h - 1}{h} &= \frac{e^{k} - 1}{k} \cdot \frac{k}{h} \\ &= \ln a \cdot \frac{e^{k} - 1}{k}. \end{aligned}

As h \to 0, k \to 0, and k \ne 0 whenever h \ne 0. So by substituting k for h inside the limit, \dfrac{e^{k} - 1}{k} \to L(e) = 1, and

L(a) = \ln a \cdot 1 = \ln a.

For a = 1, \dfrac{1^h - 1}{h} = 0 for every h, so L(1) = 0 = \ln 1. The argument also proves that L(a) exists for every base, once it exists for e.

A shorter route writes a^x = e^{x \ln a} and differentiates e raised to a function of x; that is the chain rule, which comes later. The substitution above needs only limits.

The linear exponent. By the power of a power rule, e^{kx} = \left(e^{k}\right)^x, an exponential with base b = e^{k} and \ln b = k. So its derivative is \left(e^{k}\right)^x \cdot k = k e^{kx}.

What is stated, not proved. Three facts are used without proof here. First, that a^x has a meaning for irrational x (such as \displaystyle 2^{\sqrt{3}}), filled in so that the function is continuous and the laws of exponents keep holding for all real exponents. Second, that a base with L(e) = 1 exists, which the table makes believable but does not prove. Third, that this base is the compounding number \displaystyle \lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^n. A careful proof of all three usually defines \ln x first, as an area, and builds e^x from it. Once the derivative exists, exponentials are continuous, since a function is continuous wherever it is differentiable.

Worked examples

Example 1

Reading off derivatives

Find each derivative: (a) \dfrac{d}{dx}\big[7^x\big]; (b) f'(x) for f(x) = 3e^x - 2x^4 + 6; (c) g'(t) for g(t) = (0.6)^t; (d) the derivatives of x^3, 3^x and 3^3.

Solution

(a) The base 7 is fixed and the exponent varies, so the exponential rule applies: \dfrac{d}{dx}\big[7^x\big] = 7^x \ln 7 \approx 1.9459 \cdot 7^x.

(b) The sum rule and constant multiple rule from the polynomial lesson split f into terms. The term 3e^x keeps its coefficient and e^x is its own derivative:

f'(x) = 3e^x - 8x^3.

(c) The base 0.6 is less than 1, so \ln 0.6 \approx -0.5108 is negative: g'(t) = (0.6)^t \ln 0.6. Every slope is negative, as it should be for a decreasing function.

(d) In x^3 the variable is the base: the power rule gives 3x^2. In 3^x the variable is the exponent: 3^x \ln 3. And 3^3 = 27 has no variable at all, so its derivative is 0.

Example 2

Limits that are derivatives

Evaluate (a) \displaystyle \displaystyle \lim_{h \to 0} \frac{10^h - 1}{h} and (b) \displaystyle \displaystyle \lim_{h \to 0} \frac{e^{2 + h} - e^2}{h}.

Solution

(a) This is L(10) by definition, and L(a) = \ln a, so the limit is \ln 10 \approx 2.3026. It is the slope of y = 10^x at (0, 1).

(b) This is the difference quotient of f(x) = e^x at x = 2, so the limit is f'(2). Since f'(x) = e^x, the limit is e^2 \approx 7.389.

A check: with h = 0.001, \dfrac{e^{2.001} - e^2}{0.001} \approx 7.393.

Example 3

A tangent line to y = 2^x

Find the tangent line to y = 2^x at x = 3, and where it crosses the x-axis.

Solution

The point is (3, 2^3) = (3, 8). The slope is the derivative there: \dfrac{d}{dx}\big[2^x\big] = 2^x \ln 2, so the slope is 8 \ln 2 \approx 5.545. In point-slope form, as in tangent lines,

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

Setting y = 0 gives 8 \ln 2\,(x - 3) = -8, so x - 3 = -\dfrac{1}{\ln 2} and x = 3 - \dfrac{1}{\ln 2} \approx 1.557.

The 8 canceled. At any point (c, 2^c) the slope is the height times \ln 2, so every tangent to y = 2^x crosses the x-axis the same distance \dfrac{1}{\ln 2} \approx 1.443 to the left of its point.

Example 4

Kadiatou’s radon canister

Kadiatou seals a sample of basement air in a test canister. In a model where radon loses about 16.6\% of itself each day, the concentration after t days is C(t) = 40(0.834)^t picocuries per liter. Find C'(0) and C'(5) with units, show that the rate is always the same multiple of the concentration, and find when the concentration is falling at 1 picocurie per liter per day.

Solution

The constant multiple rule keeps the 40, and the exponential rule gives

C'(t) = 40(0.834)^t \ln 0.834.

With \ln 0.834 \approx -0.18152, C'(0) = 40 \ln 0.834 \approx -7.261 picocuries per liter per day. At t = 5, (0.834)^5 \approx 0.40349, so C'(5) \approx 40(0.40349)(-0.18152) \approx -2.930 picocuries per liter per day. The concentration is falling more slowly, because there is less radon left to decay.

Dividing, \dfrac{C'(t)}{C(t)} = \ln 0.834 \approx -0.1815 for every t: whatever is in the canister, it is losing about 18.2\% of that amount per day at that instant. (This instant rate is larger than the 16.6\% actually lost over a whole day, because the amount shrinks during the day.)

Falling at 1 picocurie per liter per day means C'(t) = -1, so C(t) = \dfrac{-1}{\ln 0.834} \approx 5.509. Then

\begin{aligned} (0.834)^t &= \frac{5.509}{40}, \\ t &= \frac{\ln(5.509/40)}{\ln 0.834} \approx 10.92 \text{ days}. \end{aligned}

Example 5

Ousmane estimates from a tangent line

Ousmane wants C(2.5) for Kadiatou’s canister but uses only the value and the rate at t = 2. Find the tangent line at t = 2, use it to estimate C(2.5), and compare with the model.

Solution

C(2) = 40(0.834)^2 \approx 27.822, and C'(2) = C(2) \ln 0.834 \approx -5.050. The tangent line is

y = 27.822 - 5.050\,(t - 2).

At t = 2.5 it gives 27.822 - 5.050(0.5) \approx 25.297 picocuries per liter. The model gives C(2.5) = 40(0.834)^{2.5} \approx 25.408. The estimate is low by about 0.11: the curve bends upward, so it lies above its tangent line (the figure shows this), and a tangent-line estimate near t = 2 undershoots. Over half a day the line is still close.

The tangent reaches 0 at t = 2 + \dfrac{1}{0.18152} \approx 7.51, which is again a fixed distance \dfrac{1}{|\ln 0.834|} from its point, as in Example 3.

Figure 2The radon curve and its tangent at t = 2 (Examples 4 and 5). The tangent lies under the curve, so a tangent-line estimate near t = 2 comes out a little low.

Example 6

A linear exponent

Find f'(x) and f'(0) for f(x) = e^{3x} - 2e^{-x}.

Solution

Rewrite each term as an exponential with a fixed base by the power of a power rule: e^{3x} = \left(e^3\right)^x and e^{-x} = \left(e^{-1}\right)^x. Since \ln e^3 = 3 and \ln e^{-1} = -1,

\frac{d}{dx}\big[e^{3x}\big] = 3e^{3x} \quad \text{and} \quad \frac{d}{dx}\big[e^{-x}\big] = -e^{-x}.

With the constant multiple and difference rules, f'(x) = 3e^{3x} + 2e^{-x}, and so f'(0) = 3 + 2 = 5.

The rewrite works because the exponent is a constant times x. For an exponent such as x^2 there is no fixed base to rewrite to, and e^{x^2} waits for the chain rule.

Example 7

The tangent through the origin

Find the point on y = 3^x whose tangent line passes through the origin, and the equation of that line.

Solution

At x = c the point is (c, 3^c) and the slope is 3^c \ln 3, so the tangent is y = 3^c + 3^c \ln 3\,(x - c). It passes through (0, 0) exactly when

\begin{aligned} 0 &= 3^c + 3^c \ln 3\,(0 - c) \\ &= 3^c\,(1 - c \ln 3). \end{aligned}

Since 3^c > 0, this forces c = \dfrac{1}{\ln 3} \approx 0.9102. The height there is 3^{1/\ln 3}. Writing 3 = e^{\ln 3},

3^{1/\ln 3} = \left(e^{\ln 3}\right)^{1/\ln 3} = e^{1} = e.

So the point is \left(\dfrac{1}{\ln 3}, e\right), the slope is e \ln 3 \approx 2.986, and the line is y = (e \ln 3)\,x. The same steps for any base a > 1 give the point \left(\dfrac{1}{\ln a}, e\right): the tangent from the origin always touches at height e.

Boundaries

The base must be positive and fixed. For a \le 0, a^x is not defined for every real x (for example, (-4)^{1/2} is not real), so there is no function on an interval to differentiate. For a = 1 the function is constant and its derivative is 0.
The variable must be the exponent alone, or a constant times it. \dfrac{d}{dx}\big[a^x\big] = a^x \ln a is not the power rule, and the power rule is not this rule. A function such as x^x, with the variable in both places, or 2^{x^2}, with a nonlinear exponent, needs the chain rule or logarithmic differentiation.
The logarithm is the natural one. \dfrac{d}{dx}\big[10^x\big] = 10^x \ln 10, not 10^x \log 10 = 10^x. Only base e makes the factor 1.
Products and quotients are not covered yet. x \cdot 2^x or \dfrac{e^x}{x} cannot be split term by term; they wait for the product rule and the quotient rule.

Where this goes next

With \dfrac{d}{dx}\big[e^x\big] = e^x in hand, the derivative of \ln x follows by working backward from it, and the chain rule extends the exponential rule to e^{g(x)} for any differentiable g. Reading the rule in reverse gives antiderivatives of exponentials. Because e^x is its own derivative, every derivative at 0 equals 1, which is what builds the series for e^x. And the fact that an exponential’s rate is proportional to its size is why exponential models describe radon, bacteria and interest alike.

Exercises

Reading off derivatives

Find each derivative.

f(x) = 6^x
g(x) = (0.2)^x
y = 4e^x - 9 (give \dfrac{dy}{dx})
h(t) = 5 \cdot 3^t + t^3
p(x) = \pi^x + \pi^2 + x^2
q(x) = 2^{x + 4} (rewrite first)
r(x) = 9^{x/2} (rewrite first)
s(x) = 6e^{-x/3}

Limits that are derivatives

Evaluate each limit by recognizing a derivative.

\displaystyle \displaystyle \lim_{h \to 0} \frac{8^h - 1}{h}
\displaystyle \displaystyle \lim_{h \to 0} \frac{2^{h - 1} - \frac{1}{2}}{h}
\displaystyle \displaystyle \lim_{x \to 2} \frac{5^x - 25}{x - 2}

Tangent lines

Find the tangent line to y = e^x at x = -1.
Find the tangent line to y = 4^x at x = \tfrac{1}{2}.
Find the tangent line to y = \left(\tfrac{1}{3}\right)^x at x = -2.

Rates in the radon canister

Astou seals a second canister, with concentration C(t) = 25(0.834)^t picocuries per liter after t days.

Find C'(0) and C'(7), with units.
Measured in hours s, the model is C = 25b^s with b = (0.834)^{1/24}. Find the rate of change at s = 0 in picocuries per liter per hour.
When is the concentration falling at 2 picocuries per liter per day?
Use the tangent line at t = 3 to estimate C(3.2), and say whether the estimate is high or low.

Slopes chosen in advance

Find the point on y = 2^x where the slope is 1.
Find the base a > 0 for which y = a^x has slope -1 at x = 0.

Reasoning

Let a > 0, a \ne 1. Show that the tangent to y = a^x at any point (c, a^c) meets the x-axis at x = c - \dfrac{1}{\ln a}, so the horizontal distance does not depend on c.
Without using logarithms, show from the limit that L(a^2) = 2L(a) for a > 0, by factoring a^{2h} - 1. Check it against the table for a = 2.
Ousmane says \dfrac{d}{dx}\big[2^x\big] = x \cdot 2^{x - 1}. Explain what rule he has used and why it does not apply, and compare his formula with the table at x = 0.
Without using logarithms, show from the limit that L\!\left(\tfrac{1}{a}\right) = -L(a) for a > 0, and explain what this says about the graphs of a^x and \left(\tfrac{1}{a}\right)^x at x = 0.

Answers

f'(x) = 6^x \ln 6.
g'(x) = (0.2)^x \ln 0.2 = -(0.2)^x \ln 5. Since \ln 0.2 = -\ln 5 \approx -1.6094, the function decreases.
\dfrac{dy}{dx} = 4e^x. The constant -9 contributes 0.
h'(t) = 5 \cdot 3^t \ln 3 + 3t^2.
p'(x) = \pi^x \ln \pi + 2x. The term \pi^2 is a constant.
q'(x) = 2^{x + 4} \ln 2. Rewrite q(x) = 16 \cdot 2^x; the derivative is 16 \cdot 2^x \ln 2.
r'(x) = 3^x \ln 3. 9^{x/2} = \left(9^{1/2}\right)^x = 3^x.
s'(x) = -2e^{-x/3}. e^{-x/3} = \left(e^{-1/3}\right)^x with \ln e^{-1/3} = -\tfrac{1}{3}, and 6 \cdot \left(-\tfrac{1}{3}\right) = -2.
\ln 8 = 3 \ln 2 \approx 2.0794. It is L(8).
\tfrac{1}{2}\ln 2 \approx 0.3466. The difference quotient of 2^x at x = -1, where the derivative is 2^{-1} \ln 2.
25 \ln 5 \approx 40.236. The alternate form of the derivative of 5^x at x = 2: 5^2 \ln 5.
y = \dfrac{x + 2}{e}. Point \left(-1, e^{-1}\right), slope e^{-1}: y = e^{-1} + e^{-1}(x + 1).
y = 2 + 4 \ln 2\,\left(x - \tfrac{1}{2}\right). Point \left(\tfrac{1}{2}, 2\right), slope 4^{1/2} \ln 4 = 2 \ln 4 = 4 \ln 2 \approx 2.7726.
y = 9 - 9\ln 3\,(x + 2). Point (-2, 9), slope 9 \ln \tfrac{1}{3} = -9 \ln 3 \approx -9.888.
C'(0) \approx -4.538 and C'(7) \approx -1.274 picocuries per liter per day. C'(t) = 25(0.834)^t \ln 0.834; (0.834)^7 \approx 0.28065.
About -0.1891 picocuries per liter per hour. C'(0) = 25 \ln b = \dfrac{25 \ln 0.834}{24}, the daily rate divided by 24.
At t \approx 4.51 days. C(t) = \dfrac{-2}{\ln 0.834} \approx 11.018, so (0.834)^t \approx 0.44072 and t = \dfrac{\ln 0.44072}{\ln 0.834}.
About 13.976, low. C(3) \approx 14.502 and C'(3) \approx -2.632, so the line gives 14.502 - 0.2(2.632). The model gives C(3.2) \approx 13.985; the curve lies above its tangent.
At x \approx 0.5288, the point \left(0.5288, \tfrac{1}{\ln 2}\right) \approx (0.5288, 1.4427). 2^c \ln 2 = 1 gives 2^c = \tfrac{1}{\ln 2}, so c = \log_2 \tfrac{1}{\ln 2} = -\dfrac{\ln(\ln 2)}{\ln 2} by change of base.
a = \dfrac{1}{e}. The slope at 0 is \ln a, and \ln a = -1.
The tangent is y = a^c + a^c \ln a\,(x - c). Setting y = 0 and dividing by a^c \ln a, which is not 0, gives x - c = -\dfrac{1}{\ln a}. The a^c canceled, so the distance \dfrac{1}{|\ln a|} is the same at every point (to the left when a > 1, to the right when a < 1).
Factor the top as a difference of squares: \dfrac{a^{2h} - 1}{h} equals \dfrac{a^h - 1}{h} times (a^h + 1). The first factor tends to L(a). Also a^h - 1 is h times \dfrac{a^h - 1}{h}, so it tends to 0 \cdot L(a) = 0, and a^h + 1 \to 2. The product law gives L(a^2) = 2L(a). For a = 2: L(4) = 2L(2) \approx 1.386, which is \ln 4.
He used the power rule, which is for x^n with the variable in the base and a fixed exponent. In 2^x the exponent varies, so the power rule says nothing. At x = 0 his formula gives 0, but the table shows the slope of 2^x at 0 is about 0.693; the correct derivative gives 2^0 \ln 2 = \ln 2.
Since \left(\tfrac{1}{a}\right)^h = a^{-h}, the quotient is \dfrac{a^{-h} - 1}{h}, and factoring out -a^{-h} turns it into -a^{-h} \cdot \dfrac{a^h - 1}{h}. As h \to 0, a^{-h} = \dfrac{1}{a^h} \to 1 (as in Exercise 22) and the second factor tends to L(a), so the limit is -L(a). The graphs of a^x and \displaystyle \left(\frac{1}{a}\right)^x are mirror images across the y-axis, so at (0, 1) their slopes are opposite.