13/03/2026

#Math#Differentiation

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Exponential Models and Logarithmic Differentiation

Lesson 5AM produced the family $y = C \, e^{k t}$ as the solutions of the rate equation $y' = k \, y$ , and Lesson 5PM named $k = \ln b$ and gave the four log identities together with logarithmic differentiation. Two pieces of data fix $C$ and $k$ for any application; one piece of data fixes one of them when the other is read off the model. This recitation applies those tools to two-data-point growth, half-life, the time constant of a decay curve, mixed-rule expressions involving $e^{x}$ and $\ln x$ , and higher-derivative patterns.

Note (Log identities used below)

We use the labels from Lesson 5PM: LI is $\ln(xy) = \ln x + \ln y$ , LII is $\ln(1/x) = -\ln x$ , LIII is $\ln(x/y) = \ln x - \ln y$ , and LIV is $\ln(x^{b}) = b\ln x$ , with positive logarithm arguments.

Two Data Points Fix $C \, e^{k t}$

A measurement at one time fixes $C$ . A measurement at a second time then fixes $k$ through one application of $\ln$ .

Example 1 (A bacteria culture from two readings)

A bacteria culture grows so that $N'(t) = k \, N(t)$ for some constant $k > 0$ , with $t$ measured in hours. At time $t = 0$ there are $20\,000$ bacteria, and at time $t = 5$ there are $400\,000$ bacteria. Find $N(t)$ in closed form.

By the solutions theorem, $N(t) = C \, e^{k t}$ . The reading at $t = 0$ gives

20\,000 = N(0) = C \, e^{0} = C,

so $C = 20\,000$ .

The reading at $t = 5$ gives

20\,000 \, e^{5 k} = 400\,000, \qquad e^{5 k} = 20.

Take logarithms of both sides. Since $\ln(e^{5k}) = 5k$ , this gives

5 k = \ln 20, \qquad k = \frac{\ln 20}{5}.

Hence

N(t) = 20\,000 \, e^{(\ln 20 / 5) \, t}.

The growth rate at any moment is $k \, N(t)$ , with $k = (\ln 20)/5 \approx 0{.}599$ per hour.

Example 2 (A doubling time fixes

k

alone)

A colony of fruit flies grows so that $P(t) = P_{0} \, e^{k t}$ , with $t$ in days. The colony’s size doubles in $9$ days. Find $k$ .

The doubling condition is $P(9) = 2 \, P(0)$ . Substituting the model and dividing by $P_{0}$ ,

e^{9 k} = 2.

Take $\ln$ of both sides:

9 k = \ln 2, \qquad k = \frac{\ln 2}{9} \approx 0{.}077 \text{ per day}.

The initial population $P_{0}$ never appeared. The growth constant of an exponential model is fixed by any ratio of values at two times, not by the absolute scale.

Problem 1

A continuation of the fruit-fly example. The initial population is $P_{0} = 100$ .

Use $k = (\ln 2)/9$ to write $P(t)$ explicitly.
Compute the size of the colony at $t = 41$ days, in closed form involving a power of $2$ . Compute the rate of growth at that moment in flies per day, leaving the answer in terms of $\ln 2$ .
Find the time at which the colony first contains $800$ flies. Reduce the equation to one of the form $e^{k t} = 8$ , then solve in closed form using $\ln$ .

Problem 2

A British investor opens an account at time $t = 0$ with $P$ pounds. The balance grows continuously at a fixed rate, so $B(t) = P \, e^{r t}$ . After $4$ years the balance is £1500, and after $10$ years it is £3000.

Use the ratio $B(10) / B(4)$ to find $r$ in closed form, without computing $P$ .
Use either reading to find $P$ .
Show, using the answer to part (1) and the doubling-time logic of the fruit-fly example, that the doubling time of this account is exactly $6$ years.

Problem 3

A laboratory yeast culture in early growth phase has cell count $N(t) = N_{0} \, e^{k t}$ , with $t$ in hours. Two technicians take readings: $N(2) = 1{.}5 \times 10^{6}$ and $N(7) = 4{.}5 \times 10^{6}$ .

Find $k$ by using the ratio of the two readings, expressing $k$ in closed form involving $\ln 3$ .
Find $N_{0}$ in closed form.
Find the time at which the cell count first reaches $10^{7}$ , in closed form.

Half-Life Applications

For a decay model $M(t) = M_{0} \, e^{-\lambda t}$ with $\lambda > 0$ , the half-life formula $T_{1/2} = (\ln 2)/\lambda$ from Lesson 5PM converts every decay constant into a half-life and back. The fractional rate of decay $|M'(t)|/M(t) = \lambda$ is constant in $t$ , so every interval of length $T_{1/2}$ removes half of whatever is present at its start.

Example 3 (Strontium-90)

Strontium- $90$ decays at rate $-\lambda \, M(t)$ with $\lambda = 0{.}0244$ per year. Find the half-life $T_{1/2}$ , and the fraction of the original mass remaining after $100$ years.

By the half-life formula,

T_{1/2} = \frac{\ln 2}{\lambda} = \frac{\ln 2}{0{.}0244} \approx 28{.}4 \text{ years}.

The mass remaining after $100$ years is

\frac{M(100)}{M_{0}} = e^{-100 \lambda} = e^{-2{.}44} \approx 0{.}0872,

so about $8{.}72\%$ of the strontium- $90$ remains. Equivalently, $100 / T_{1/2} \approx 3{.}52$ , so the sample has passed through just over three and a half half-lives.

Example 4 (Iodine-131 in dairy hay)

Iodine- $131$ has half-life $8$ days. Hay collected after a nuclear test contains $10$ times the maximum allowable level of iodine- $131$ for use as cattle fodder. How many days must the hay be stored before it is safe?

The decay constant is $\lambda = (\ln 2)/8$ per day, and the iodine content at time $t$ is $M(t) = M_{0} \, e^{-\lambda t}$ . The first safe time occurs at the threshold $M(t) = M_{0}/10$ , so we solve

e^{-\lambda t} = \frac{1}{10}.

Take logarithms of both sides. Since $\ln(e^{-\lambda t}) = -\lambda t$ and LII gives $\ln(1/10) = -\ln 10$ , we get

-\lambda t = -\ln 10, \qquad t = \frac{\ln 10}{\lambda} = \frac{8 \ln 10}{\ln 2}.

Numerically, $t \approx 8 \cdot 2{.}303 / 0{.}693 \approx 26{.}6$ days. The hay must be stored at least $27$ days.

Remark

The bound ” $M(t) \leq M_{0}/10$ for all later $t$ ” is automatic once $M$ first reaches $M_{0}/10$ , because $M$ is strictly decreasing. The strict monotonicity of $e^{-\lambda t}$ for $\lambda > 0$ is one application of the formula for $e^{kx}$ with $k = -\lambda < 0$ .

Problem 4

Carbon- $14$ has half-life $5730$ years. A parchment fragment contains $80\%$ of the carbon- $14$ that a comparable living organism would have today. Estimate the age of the parchment in closed form, then compute it numerically to the nearest year using $\ln 0{.}8 \approx -0{.}2231$ .

Set $M(t) = M_{0} \, e^{-\lambda t}$ with $\lambda = (\ln 2)/5730$ , and solve $M(t)/M_{0} = 0{.}8$ for $t$ .

Problem 5

A radioactive sample drops from $80$ grams to $5$ grams over a $24$ -hour interval, following $M(t) = M_{0} \, e^{-\lambda t}$ .

Use the ratio $M(24)/M(0) = 5/80$ to find $\lambda$ in closed form involving $\ln 2$ .
Find the half-life $T_{1/2}$ of the isotope. Confirm $24$ hours equals $4 T_{1/2}$ exactly.
Predict the mass at $t = 36$ hours, in closed form, by computing $M_{0} \cdot 2^{-36/T_{1/2}}$ .

Problem 6

Two radioactive isotopes A and B have decay constants $\lambda_{A}$ and $\lambda_{B} = 2 \lambda_{A}$ . Both samples have the same initial mass $M_{0}$ .

Express the ratio $M_{B}(t) / M_{A}(t)$ in closed form using the formula for $e^{k x}$ , and show that the ratio is itself an exponential decay $e^{-\lambda_{A} t}$ .
Show that the time required for B’s mass to fall to half of A’s mass is the half-life $T_{A,1/2}$ of A.
Compute the time at which $M_{B}(t)$ equals one-eighth of $M_{A}(t)$ , in closed form involving $T_{A,1/2}$ .

The Time Constant of a Decay Curve

A common shorthand for an exponential decay is the time constant. For $y = C \, e^{-\lambda t}$ with $\lambda > 0$ , the tangent line at $t = 0$ has slope $-C \lambda$ and passes through $(0, C)$ , so its equation is

y = C (1 - \lambda t).

The tangent meets the $t$ -axis when $y = 0$ , that is, at $t = 1/\lambda$ .

Definition 1 (Time constant)

For an exponential decay $y = C \, e^{-\lambda t}$ with $\lambda > 0$ , the time constant is

\tau = \frac{1}{\lambda}.

$A decay curve y = C e^{-λt} drawn solid. The tangent line at t = 0 is dashed and meets the t-axis at t = τ = 1/λ. The actual curve at t = τ sits at C/e, well above the zero of the tangent.$

The curve at $t = \tau$ has height $C \, e^{-\lambda \tau} = C \, e^{-1} = C/e \approx 0{.}368 \, C$ , while the tangent line is already at zero. Thus the time constant is the tangent-line intercept, not the actual time at which the decay reaches zero.

The relationship between $\tau$ and the half-life is one substitution:

T_{1/2} = \frac{\ln 2}{\lambda} = \tau \ln 2.

The half-life is shorter than the time constant by the factor $\ln 2 \approx 0{.}693$ .

Example 5 (Sales after advertising stops)

A clothing retailer’s monthly sales fall along the curve $S(t) = S_{0} \, e^{-\lambda t}$ in the months following the end of a promotional campaign. In the last month of advertising, sales were $S_{0} = 5000$ units. Six months after the campaign ended, sales had fallen to $2000$ units. Find the time constant and the half-life of the sales decay.

The reading at $t = 6$ gives

5000 \, e^{-6 \lambda} = 2000, \qquad e^{-6 \lambda} = 0{.}4.

Take logarithms of both sides; LII gives $\ln(0{.}4) = -\ln 2{.}5$ :

-6 \lambda = -\ln 2{.}5, \qquad \lambda = \frac{\ln 2{.}5}{6} \approx 0{.}153 \text{ per month}.

The time constant is

\tau = \frac{1}{\lambda} = \frac{6}{\ln 2{.}5} \approx 6{.}55 \text{ months},

and the half-life is $\tau \ln 2 = (6 \ln 2)/(\ln 2{.}5) \approx 4{.}54$ months. By month $\tau \approx 6{.}55$ , sales have fallen to $S_{0}/e \approx 1840$ , the standard ” $37\%$ remaining” reading on a decay tangent.

Problem 7

A drug’s bloodstream concentration after intravenous injection follows $C(t) = C_{0} \, e^{-\lambda t}$ , with $t$ in hours. The time constant for the clearance is $\tau = 4$ hours.

State $\lambda$ in closed form.
Compute the half-life of the clearance, in closed form involving $\ln 2$ .
Compute the fraction of the initial dose remaining at $t = \tau$ , $t = 2 \tau$ , and $t = 3 \tau$ . Express each in the form $e^{-n}$ for an integer $n$ .

Problem 8

A capacitor in an electronics circuit discharges so that its voltage satisfies $V(t) = V_{0} \, e^{-t / \tau}$ for a fixed positive constant $\tau$ . Show, using only the formula for $e^{k x}$ , that the fractional rate of discharge $|V'(t)|/V(t)$ is constant in $t$ and equals $1/\tau$ . Hence any time interval of length $\tau$ multiplies the voltage by $1/e$ , regardless of where the interval starts.

Mixed-Rule Applications

The two functions below are useful test cases for the new rules. Each is one chain rule away from the tools of Lessons 5AM and 5PM.

Example 6 (A Gaussian-shaped curve)

The function $f(x) = e^{-x^{2}}$ is a Gaussian-shaped curve. Find $f'(x)$ , locate every critical number, and find every inflection point.

By the chain rule for exponentials with $g(x) = -x^{2}$ ,

f'(x) = e^{-x^{2}} \cdot (-2 x) = -2 x \, e^{-x^{2}}.

The factor $e^{-x^{2}}$ is strictly positive, so $f'(x) = 0$ exactly when $x = 0$ . The first derivative test applies: $f'$ is positive for $x < 0$ and negative for $x > 0$ , so there is a relative maximum at $x = 0$ . The maximum point is $(0, 1)$ , and the maximum value is $1$ .

For inflection points, differentiate again. By the product rule on $-2 x$ and $e^{-x^{2}}$ ,

f''(x) = -2 \, e^{-x^{2}} + (-2 x) \cdot (-2 x) \, e^{-x^{2}} = (4 x^{2} - 2) \, e^{-x^{2}}.

Since $e^{-x^{2}} > 0$ always, $f''(x) = 0$ exactly when $4 x^{2} - 2 = 0$ , that is, when $x = \pm 1/\sqrt{2}$ . The sign of $4 x^{2} - 2$ flips at each value, so by the inflection criterion the inflection points are $\left(-1/\sqrt{2}, e^{-1/2}\right)$ and $\left(1/\sqrt{2}, e^{-1/2}\right)$ .

$The Gaussian-shaped curve y = e^{-x²} drawn from x = -2.5 to x = 2.5, peaking at the marked point (0, 1) and decaying symmetrically to both sides. Two inflection points are marked at x = ±1/√2 with their heights, slightly above 0.6.$

The calculus locates the two bending changes with one application each of the chain rule and the product rule.

Example 7 (The softplus)

The function $g(x) = \ln(1 + e^{x})$ is often called the softplus function. Show that $0 < g'(x) < 1$ for every real $x$ , and identify the asymptotic behaviour of $y = g(x)$ as $x \to -\infty$ and as $x \to \infty$ .

By the chain rule for $\ln$ with $h(x) = 1 + e^{x}$ , $h'(x) = e^{x}$ ,

g'(x) = \frac{e^{x}}{1 + e^{x}}.

Both numerator and denominator are positive for every real $x$ , and the numerator is strictly less than the denominator, so $0 < g'(x) < 1$ .

For the asymptotic behaviour, factor $e^{x}$ out of the argument of the logarithm:

g(x) = \ln(1 + e^{x}) = \ln \!\left( e^{x} (e^{-x} + 1) \right) = \ln(e^{x}) + \ln(e^{-x} + 1) = x + \ln(1 + e^{-x}).

As $x \to \infty$ , $e^{-x} \to 0$ (Lesson 5AM, decreasing-curve case of $e^{k x}$ with $k = -1$ ), so $\ln(1 + e^{-x}) \to \ln 1 = 0$ . Therefore $g(x) - x \to 0$ , so the graph approaches the line $y = x$ on the right. As $x \to -\infty$ , $e^{x} \to 0$ , so $g(x) \to \ln 1 = 0$ ; the horizontal line $y = 0$ is the asymptote on the left.

The softplus curve y = ln(1 + e^x) from x = -5 to x = 5. The curve flattens to y = 0 on the left and rises along the line y = x on the right. Both limiting lines are drawn dotted.

The softplus is a smooth replacement for the kinked piecewise function that equals $0$ for $x \leq 0$ and equals $x$ for $x > 0$ : it has the same limiting shape, but its derivative exists everywhere.

Problem 9

Differentiate. Each is one or two applications of the rules from Lessons 5AM and 5PM.

$y = e^{x} (x^{2} - 3 x + 1)$ , then factor $y'$ .
$y = \dfrac{e^{2 x}}{1 + e^{x}}$ , simplifying $y'$ as a single fraction.
$y = \ln \!\left( \dfrac{e^{x} + 1}{e^{x} - 1} \right)$ on $x > 0$ . Split using LIII before differentiating.
$y = (1 + e^{x})^{1/2}$ .
$y = \ln \!\left( \sqrt{x^{2} + 4} \cdot e^{x} \right)$ . Simplify using LI and LIV first.

Problem 10

The sigmoid function is

\sigma(x) = \frac{1}{1 + e^{-x}},

Show by direct differentiation that

\sigma'(x) = \sigma(x) \, \bigl(1 - \sigma(x)\bigr).

This identity expresses the derivative entirely in terms of $\sigma(x)$ .

Problem 11

For each of the following, find every relative extreme point on the indicated domain and classify it using either the first or the second derivative test.

$f(x) = x \, e^{-x}$ on the real line.
$f(x) = (\ln x)^{2}$ on $x > 0$ .
$f(x) = x^{2} \, e^{-x^{2}}$ on the real line.
$f(x) = \dfrac{e^{x}}{x}$ on $x > 0$ .

Higher Derivatives

The derivative formulas $\dfrac{d}{dx}(e^{k x}) = k \, e^{k x}$ and $\dfrac{d}{dx}(\ln x) = 1/x$ are stable enough under repeated differentiation to admit closed-form $n$ th derivatives.

Recitation 2 named the first two derivatives of a position function $s(t)$ : the velocity $v(t) = s'(t)$ and the acceleration $a(t) = s''(t)$ . The pattern continues. The third derivative $s'''(t)$ is the rate at which acceleration itself changes. For our purposes, higher derivatives are simply the functions obtained by differentiating more than twice.

The notation

\frac{d^{n}}{dx^{n}} f(x)

means applying $\dfrac{d}{dx}$ to $f$ a total of $n$ times. Thus $n = 1$ gives $f'$ , $n = 2$ gives $f''$ , and $n = 3$ gives $f'''$ .

Example 8 (Higher derivatives of

x \, e^{x}

)

Find a closed form for the $n$ th derivative of $f(x) = x \, e^{x}$ .

By the product rule,

f'(x) = 1 \cdot e^{x} + x \cdot e^{x} = (x + 1) \, e^{x}.

Differentiating again,

f''(x) = e^{x} + (x + 1) \, e^{x} = (x + 2) \, e^{x},

and once more,

f'''(x) = e^{x} + (x + 2) \, e^{x} = (x + 3) \, e^{x}.

Each application of the product rule introduces a new $e^{x}$ from the derivative of the polynomial factor and leaves the polynomial part unchanged, so the constant term grows by $1$ each round. Differentiating $(x + n) \, e^{x}$ by the product rule produces

\frac{d}{dx} \bigl[(x + n) \, e^{x}\bigr] = e^{x} + (x + n) \, e^{x} = (x + n + 1) \, e^{x},

so once the formula holds at level $n$ , the same product rule applied once more delivers it at level $n + 1$ . The first three derivatives above start the chain at $n = 1, 2, 3$ , and every subsequent level continues by the same one-step calculation:

\frac{d^{n}}{d x^{n}} (x \, e^{x}) = (x + n) \, e^{x}.

Problem 12

For $g(x) = e^{k x}$ with constant $k$ , compute $g'$ , $g''$ , $g'''$ , and $g^{(4)}$ directly, identify the pattern, and write down the closed form for the $n$ th derivative

\frac{d^{n}}{d x^{n}} (e^{k x}) = k^{n} \, e^{k x}

together with a one-line argument that differentiating $k^{n} \, e^{k x}$ once more produces $k^{n + 1} \, e^{k x}$ , so each level forces the next. Use the formula to compute the third and fourth derivatives of $y = 5 \, e^{-2 x}$ at $x = 0$ .

Problem 13

Find the $n$ th derivative of $f(x) = \ln x$ on $x > 0$ , for $n \geq 1$ . Compute the first four derivatives, identify the pattern, and express the closed form using $(-1)^{n - 1}$ , a factorial of $n - 1$ , and a power of $x$ . Here $(n - 1)!$ means the product $1 \cdot 2 \cdots (n - 1)$ , with $0! = 1$ .

Problem 14

For $f(x) = (x^{2} + x) \, e^{x}$ , compute $f'(x)$ , $f''(x)$ , and $f'''(x)$ , factoring the polynomial part each time. Predict $f^{(4)}(x)$ from the pattern, then differentiate to verify.

Exercises

Exercise 1

A bacteria culture grows so that $N(0) = 1000$ and $N(3) = 8000$ , with $N(t) = N_{0} \, e^{k t}$ .

Find $k$ in closed form involving $\ln 2$ .
Find the doubling time of the culture, in closed form. Verify it equals $1$ hour.
Predict $N(10)$ in closed form, leaving the answer as a power of $2$ multiplied by $1000$ .

Exercise 2

A radioactive isotope has half-life $T_{1/2}$ years. Working only from $M(t) = M_{0} \, e^{-(\ln 2 / T_{1/2}) t}$ , show that:

The fraction of the original mass remaining after $n T_{1/2}$ years is exactly $2^{-n}$ , for every positive integer $n$ .
The time required for the mass to decay to $1\%$ of its original value is $T_{1/2} \cdot \dfrac{\ln 100}{\ln 2}$ .
The fractional rate of decay $|M'(t)|/M(t)$ is the constant $(\ln 2)/T_{1/2}$ at every $t$ .

Exercise 3

The function $f(x) = x \, e^{-x^{2}/2}$ is a line multiplied by a Gaussian-shaped factor.

Find every relative extreme point of $f$ on the real line, and classify each.
Determine the maximum value of $|f|$ and the points at which it is attained.
Find every inflection point of $f$ , in closed form.

Exercise 4

A British investor places $P$ pounds in an account earning continuous interest at rate $r$ per year, so the balance is $B(t) = P \, e^{r t}$ .

Determine $r$ if the balance doubles in $12$ years; leave the answer in closed form.
Using that $r$ , find the time at which the balance first reaches $10P$ pounds, in closed form involving $\ln 10$ and $\ln 2$ .
Compute the rate of growth $B'(t)$ at the moment the balance first reaches $2P$ pounds, in closed form involving $r$ and $P$ .

Exercise 5

For the softplus $g(x) = \ln(1 + e^{x})$ :

Compute $g''(x)$ and show that $g$ is concave up everywhere on the real line.
Use the rewrite $g(x) = x + \ln(1 + e^{-x})$ from the worked example to verify directly that $g(x) - x \to 0$ as $x \to \infty$ .
Find every $x$ at which $g'(x) = 1/2$ , in closed form. Compute $g$ at that point.

Exercise 6

A drug’s bloodstream concentration is modelled by

C(t) = C_{0} \, t \, e^{-\lambda t}, \qquad t \geq 0, \quad \lambda > 0,

the same delay-then-decay family that appeared as a problem in Lesson 5AM.

Find $C'(t)$ and locate the time at which the concentration peaks.
Express the peak concentration in closed form in terms of $C_{0}$ , $\lambda$ , and $e$ .
Show that the peak time equals the time constant of the underlying decay $e^{-\lambda t}$ . Interpret in one sentence.

Exercise 7

Let $f(x) = x^{x}$ on $x > 0$ , and let $g(x) = (\ln x)^{x}$ on $x > 1$ . Differentiate each by logarithmic differentiation, then locate every critical number of $f$ and of $g$ in closed form. State, with reasons drawn from this recitation and Lesson 5PM, whether each critical number is a relative minimum, maximum, or neither.

Exercise 8

Find the $n$ th derivative of $f(x) = (a x + b) \, e^{c x}$ in closed form for constants $a, b, c$ and $n \geq 1$ . Your formula should also work when $c = 0$ . Verify it by specialising to $a = 1, b = 0, c = 1$ to recover $\dfrac{d^{n}}{dx^{n}}(x \, e^{x}) = (x + n) e^{x}$ , and to $a = 0, b = 1, c = k$ to recover $\dfrac{d^{n}}{dx^{n}}(e^{k x}) = k^{n} e^{k x}$ .