Corrected exercises: exponential and logarithmic functions (201-103)

$f'(x)=3e^{1.5x}\times 1.5=4.5\,e^{1.5x}$. At $x=0.4$ the exponent is $0.6$: $f'(0.4)=4.5\,e^{0.6}\approx 8.200$. Pitfall: forgetting the $1.5$ and answering $3e^{0.6}\approx 5.466$.

Constant base, variable exponent: $g'(x)=4^{3x}\ln 4\times 3=3\ln 4\cdot 4^{3x}$. At $x=0.5$, $4^{1.5}=8$, so $g'(0.5)=24\ln 4\approx 33.271$. Pitfall: the power rule, $3x\cdot 4^{3x-1}$, which holds only for a constant exponent.

The constant $150$ disappears: $Q'(t)=-90\,e^{-0.35t}\times(-0.35)=31.5\,e^{-0.35t}$, the two minus signs giving a plus. $Q'(2)=31.5\,e^{-0.7}\approx 15.642$.

$h'(x)=(2x-3)\,e^{x^{2}-3x}$: copy the exponential and multiply by the derivative of the exponent. At $x=0.5$: $2x-3=-2$ and $x^{2}-3x=-1.25$, so $h'(0.5)=-2e^{-1.25}\approx -0.573$.

Product: $f'(x)=1\cdot e^{-2x}+x\cdot(-2)e^{-2x}=(1-2x)\,e^{-2x}$. At $x=1$: $f'(1)=-e^{-2}\approx -0.135$. Pitfall: differentiating $e^{-2x}$ as $e^{-2x}$, without the $-2$, which gives $2e^{-2}$.

Quotient with $u=e^{x}-1$, $v=e^{x}+1$ and $u'=v'=e^{x}$: $g'(x)=\frac{e^{x}(e^{x}+1)-(e^{x}-1)e^{x}}{(e^{x}+1)^{2}}=\frac{2e^{x}}{(e^{x}+1)^{2}}$. At $x=1$: $g'(1)=\frac{2e}{(e+1)^{2}}\approx 0.393$.

$h'(x)=\frac{2x+4}{x^{2}+4x+1}$: the derivative of the inside over the inside. At $x=2$: $h'(2)=\frac{8}{13}\approx 0.615$. Pitfall: writing $\frac{1}{x^{2}+4x+1}$, without the numerator $u'$.

$(\log_{2}u)'=\frac{u'}{u\ln 2}$, so $f'(x)=\frac{3}{(3x+1)\ln 2}$. At $x=1$: $f'(1)=\frac{3}{4\ln 2}\approx 1.082$. Pitfall: forgetting the $\ln 2$, which comes from $\log_{2}u=\frac{\ln u}{\ln 2}$.

Expand first: $g(x)=\ln(2x+1)-\ln(x-1)$, valid at $x=2$ since both arguments are positive there. So $g'(x)=\frac{2}{2x+1}-\frac{1}{x-1}$ and $g'(2)=\frac{2}{5}-1=-\frac{3}{5}=-0.6$. Pitfall: $\ln\frac{a}{b}=\ln a-\ln b$, a DIFFERENCE, not a quotient of logarithms.

Product: $h'(x)=3x^{2}\ln x+x^{3}\cdot\frac{1}{x}=3x^{2}\ln x+x^{2}$. At $x=2$: $h'(2)=12\ln 2+4\approx 12.318$. The $x^{3}\cdot\frac{1}{x}$ simplifies to $x^{2}$ before evaluating.

Quotient: $f'(x)=\frac{\frac{1}{x}\cdot x^{2}-\ln x\cdot 2x}{x^{4}}=\frac{x-2x\ln x}{x^{4}}=\frac{1-2\ln x}{x^{3}}$. At $x=0.5$, $\ln 0.5=-\ln 2$ and $x^{3}=\frac{1}{8}$: $f'(0.5)=8(1+2\ln 2)\approx 19.090$. Pitfall: the sign of $\ln 0.5$, which is NEGATIVE.

Power first, $g(x)=(\ln x)^{3}$: $g'(x)=3(\ln x)^{2}\cdot\frac{1}{x}$. At $x=e^{2}$, $\ln x=2$: $g'(e^{2})=\frac{3\times 4}{e^{2}}=\frac{12}{e^{2}}\approx 1.624$. Pitfall: confusing $(\ln x)^{3}$ with $\ln(x^{3})=3\ln x$, whose derivative is $\frac{3}{x}$.

$f'(x)=e^{x}$: the exponential with base $e$ is its own derivative, and it is the only nonzero function that is. $g'(x)=2^{x}\ln 2$: for any base $a>0$, $\left(a^{x}\right)'=a^{x}\ln a$, the factor $\ln a$ being the trace of the base. $h'(x)=0.05\,e^{0.05x}$ by the chain rule, the $0.05$ coming from the derivative of the exponent.

$f'(0)=e^{0}=1$. $g'(0)=2^{0}\ln 2=\ln 2\approx 0.6931$. Both curves pass through the point $(0, 1)$, but they do not have the same slope there: that of $e^{x}$ is exactly 1, that of $2^{x}$ about 0.69.

$e$ is the only base whose curve has a slope of EXACTLY 1 at the point $(0, 1)$, that is, the only base $a$ such that $\ln a=1$. Everything else follows from this: since the factor $\ln a$ equals 1, the derivative of $e^{x}$ is $e^{x}$ itself. So this is not a notational convention but a geometric property, and it is why all growth models are written with base $e$ even when the phenomenon doubles or grows tenfold.

$k'(x)=5\times 3^{x}\ln 3$. At $x=2$: $k'(2)=5\times 9\times \ln 3=45\times 1.0986\approx 49.44$. The constant 5 passes through the differentiation unchanged, whereas the base 3 leaves its logarithm: two constants, two roles, and mixing them up is the error in part e) of exercise 8.

In every case, the derivative is PROPORTIONAL to the function itself: $f'=1\cdot f$, $g'=(\ln 2)\,g$, $h'=0.05\,h$, $k'=(\ln 3)\,k$. An exponential is exactly a function whose growth rate is proportional to what is already there, which makes it unavoidable whenever a phenomenon feeds on itself: capital that earns interest, a population that produces births, a rumour spread by those who already know it.

The logarithm requires $3x-12>0$, so $x>4$: the domain is $(4, \infty)$. By the chain rule, $f'(x)=\dfrac{3}{3x-12}$, which simplifies to $\dfrac{1}{x-4}$. The simplification is legitimate on the domain, and it does not change it: $\dfrac{1}{x-4}$ would also exist for $x<4$, but the original function does not.

$x^{2}>0$ for every $x\ne 0$, so the domain is $\mathbb{R}\setminus\{0\}$. The derivative is $g'(x)=\dfrac{2x}{x^{2}}=\dfrac{2}{x}$, valid on this whole domain, negative values included. The function $2\ln x$, on the other hand, exists only on $(0, \infty)$ and its derivative $\dfrac{2}{x}$ is valid only there. The two expressions agree on the positive numbers and differ in their domains, which is exactly the pitfall of question e) of “Five statements to correct” in the first chapter: $\ln(x^{2})=2\ln|x|$.

Product rule on the first term: $\left(x\ln x\right)'=1\times\ln x+x\times\dfrac{1}{x}=\ln x+1$. Then subtract the derivative of $x$, which is 1: $h'(x)=\ln x+1-1=\ln x$. The function $x\ln x-x$ is therefore an antiderivative of the logarithm, a result that will come back in integral calculus, and obtaining it here takes nothing more than the product rule.

Every logarithm reduces to the natural logarithm: $\log_{10}x=\dfrac{\ln x}{\ln 10}$, so $p'(x)=\dfrac{1}{x\ln 10}$. At $x=10$: $p'(10)=\dfrac{1}{10\times 2.3026}\approx 0.0434$. The number is small, and that is the whole point of a logarithmic scale: multiplying the quantity by 10 raises the indicator by only one unit.

Because the derivative forgets the domain. $\dfrac{1}{x-4}$ no longer carries any trace of the condition $x>4$, and nothing in this expression prevents you from evaluating it at $x=0$ to get $-0.25$, a perfectly wrong number. The same pitfall applies to $\dfrac{2}{x}$ against $\ln(x^{2})$. In the social sciences the consequence is concrete: a model in $\ln$ only makes sense for strictly positive values, such as incomes, headcounts and prices, and extending the calculation below that produces a figure the calculator accepts and reality rejects.

The outer layer is the exponential, the inner one is $-0.4t^{2}$, whose derivative is $-0.8t$. So $u'(t)=-0.8t\,e^{-0.4t^{2}}$. The sign follows that of $-0.8t$, since the exponential is always positive: $u$ increases for $t<0$, decreases for $t>0$, and its maximum is at $t=0$. This is the bell curve, which you meet again in statistics.

$v'(t)=\dfrac{2t}{t^{2}+9}$. The domain is all of $\mathbb{R}$, since $t^{2}+9\ge 9>0$ for every $t$: this is one of the rare logarithms whose domain removes nothing. Checking takes three seconds and avoids writing a nonexistent restriction, which costs as many marks as a forgotten one.

Quotient rule with $u=e^{0.2t}$ and $v=t+5$: $w'(t)=\dfrac{0.2e^{0.2t}(t+5)-e^{0.2t}}{(t+5)^{2}}$. Factor out $e^{0.2t}$: $w'(t)=\dfrac{e^{0.2t}\bigl(0.2(t+5)-1\bigr)}{(t+5)^{2}}=\dfrac{e^{0.2t}\,(0.2t+1-1)}{(t+5)^{2}}=\dfrac{0.2t\,e^{0.2t}}{(t+5)^{2}}$. Factoring out the exponential is the reflex to build: it never vanishes, so it is ALWAYS the other factor that carries the zeros and the signs.

$w'(5)=\dfrac{0.2\times 5\times e^{1}}{100}=\dfrac{2.7183}{100}\approx 0.0272$.

Since the exponential never vanishes and the denominator is a square, the only zero of $w'$ comes from the factor $0.2t$, so $t=0$. At that point, $w(0)=\dfrac{e^{0}}{5}=\dfrac{1}{5}=0.2$. Since $w'$ has the sign of $t$ on the domain $t>-5$, the function decreases then increases: $t=0$ is a MINIMUM, not the maximum one expects by reflex from an expression containing an increasing exponential.

By the chain rule, $P'(t)=8{,}000\times 0.023\,e^{0.023t}=0.023\times 8{,}000\,e^{0.023t}=0.023\,P(t)$. The derivative is therefore proportional to the population itself, with the coefficient $0.023$ that was already in the exponent. Nothing was done except applying the chain rule, and we obtain the equation that defines every exponential growth.

The relative rate is $\dfrac{P'(t)}{P(t)}\times 100=0.023\times 100=2.3$% per year, and this number does not depend on $t$: it is CONSTANT. The parameter placed in the exponent of a model with base $e$ is therefore directly the relative rate, which explains why demographers write their models this way rather than in base 1.023.

$P(0)=8{,}000$ inhabitants. $P(30)=8{,}000\,e^{0.69}=8{,}000\times 1.9937\approx 15{,}950$ inhabitants. In thirty years, the population has almost doubled, which the next exercise will explain in one line.

$P'(0)=0.023\times 8{,}000=184$ inhabitants per year. $P'(30)=0.023\times 15{,}950\approx 367$ inhabitants per year. The ABSOLUTE rate has doubled along with the population, while the RELATIVE rate has not moved by a hundredth. This is the property that makes the exponential so deceptive to the eye: growth that seems to accelerate does not accelerate at all, in proportion.

LINEAR model: the absolute growth is constant, 1,200 inhabitants per year, and the relative rate decreases toward zero because the base keeps growing. EXPONENTIAL model: the relative rate is constant, 2.3% per year, and the absolute growth increases without bound because it applies to a growing base. The choice between the two is not a matter of taste: look at the data, and if the successive DIFFERENCES are roughly constant the model is linear, while if the successive QUOTIENTS are, the model is exponential.

We look for $T$ such that $P(T)=2P(0)$, so $e^{0.023T}=2$. Taking the natural logarithm of both sides: $0.023\,T=\ln 2$, so $T=\dfrac{\ln 2}{0.023}=\dfrac{0.69315}{0.023}\approx 30.14$ years. Note that $T$ does NOT depend on the initial population: the 8,000 cancels. A doubling time is a property of the rate, not of the size.

The rule gives $\dfrac{70}{2.3}\approx 30.43$ years, against 30.14 by the exact calculation, an error of three tenths of a year out of thirty. The number 70 comes from $\ln 2\approx 0.693$: since $T=\dfrac{\ln 2}{r}$ and the rate in percent is $100r$, we have $T=\dfrac{100\ln 2}{100r}=\dfrac{69.3}{\text{rate in percent}}$. We round 69.3 up to 70 because 70 divides easily in your head by 2, 5, 7, 10 and 14, which 69.3 does not.

The model is $Q(t)=Q_{0}e^{-0.018t}$, the loss being expressed by a negative exponent. We look for $e^{-0.018t}=0.5$, so $-0.018t=\ln 0.5=-\ln 2$, and $t=\dfrac{\ln 2}{0.018}\approx 38.51$ years. The rule of 70 gives $\dfrac{70}{1.8}\approx 38.89$, still very close. The same formula therefore serves for doubling and for half-life: only the sign of the exponent changes, and it disappears at the quotient.

Exact calculation: $T=\dfrac{\ln 2}{0.06}\approx 11.55$ years. Rule of 70: $\dfrac{70}{6}\approx 11.67$ years. The gap is about six weeks over eleven and a half years, which is irrelevant for a savings decision.

Here the compounding is ANNUAL, so the capital is $C_{0}\times 1.2^{t}$ and not $C_{0}e^{0.2t}$. Doubling requires $1.2^{t}=2$, so $t=\dfrac{\ln 2}{\ln 1.2}=\dfrac{0.69315}{0.18232}\approx 3.80$ years, whereas the rule announces $\dfrac{70}{20}=3.5$ years, an error of 8%. Two causes add up. First, the rule rests on the approximation $\ln(1+r)\approx r$, excellent for small $r$ and mediocre for $r=0.2$, where $\ln 1.2$ is only 0.182 instead of 0.2. Second, interest paid once a year earns less than interest paid continuously. The rule of 70 is therefore a tool for small rates, which covers most of demography and macroeconomics, but not double-digit returns.