Corrected exercises: derivatives of logarithms, logarithmic differentiation, growth and decay (MATH 140)

a) Complete the square: $x^2 + 4x + 5 = (x + 2)^2 + 1 \ge 1 > 0$ for every real $x$, so the logarithm is always defined and the domain is $\mathbb{R}$. Chain rule with inner function $u = x^2 + 4x + 5$, $u' = 2x + 4$: $f'(x) = \frac{u'}{u} = \frac{2x + 4}{x^2 + 4x + 5}$. The denominator never vanishes, so $f'(x) = 0$ exactly when $2x + 4 = 0$: the tangent is horizontal at $x = -2$, where $f(-2) = \ln 1 = 0$. The frequent loss is to write $\frac{1}{x^2 + 4x + 5}$: the derivative of $\ln$ is one over the INSIDE, times the derivative of the inside.

b) The argument must be positive: $3x - 1 > 0$, so the domain is $x > \frac{1}{3}$. Change of base first: $g(x) = \frac{\ln(3x - 1)}{\ln 2}$, and $\frac{1}{\ln 2}$ is a CONSTANT factor. Chain rule with $u = 3x - 1$: $g'(x) = \frac{1}{\ln 2} \cdot \frac{3}{3x - 1} = \frac{3}{(3x - 1)\ln 2}$. At $x = 3$: $g'(3) = \frac{3}{8 \ln 2}$. Leave it exact; with $\ln 2 \approx 0.69$ it is a little above $\frac{1}{2}$, a sanity check and not the answer. Forgetting $\ln 2$ costs the mark: the base of the logarithm only survives in that constant.

c) Two conditions stack: $\ln x$ needs $x > 0$, and the outer $\ln$ needs $\ln x > 0$, that is $x > 1$. Domain: $(1, \infty)$. Chain rule with inner function $u = \ln x$, $u' = \frac{1}{x}$: $h'(x) = \frac{1/x}{\ln x} = \frac{1}{x \ln x}$. At $x = e$: $h'(e) = \frac{1}{e \cdot 1} = \frac{1}{e}$. Note that the formula $\frac{1}{x \ln x}$ also makes sense at $x = \frac{1}{2}$, where $h$ does not exist: the domain is read on $h$, never on $h'$.

d) The rule $\frac{d}{dx}\ln|u| = \frac{u'}{u}$ holds wherever $u \ne 0$, whatever the sign of $u$. With $u = \cos x$, $u' = -\sin x$: $k'(x) = \frac{-\sin x}{\cos x} = -\tan x$, valid wherever $\cos x \ne 0$, that is for $x \ne \frac{\pi}{2} + n\pi$. The absolute value is what makes $k$ defined on all these intervals, including those where $\cos x < 0$; without it, $\ln(\cos x)$ would exist only where $\cos x > 0$, and the same formula $-\tan x$ would hold there.

e) For $x < 0$, $|x| = -x$, so $\ln|x| = \ln(-x)$. Chain rule with $u = -x$, $u' = -1$: $\frac{d}{dx}\ln(-x) = \frac{-1}{-x} = \frac{1}{x}$. Together with the case $x > 0$, $\frac{d}{dx}\ln|x| = \frac{1}{x}$ for every $x \ne 0$. On the figure, the tangent at $x = 2$ has slope $\frac{1}{2}$, and at $x = -2$ slope $\frac{1}{-2} = -\frac{1}{2}$: the graph of $\ln|x|$ is symmetric about the $y$ axis, so the slopes are opposite, and the left branch decreases as $x$ increases toward $0$. A negative derivative on the left branch is exactly what $\frac{1}{x}$ predicts.

a) For $x > 0$ every factor is positive, so the laws apply without absolute values: $f(x) = 3 \ln x + 4 \ln(2x + 1) - \frac{1}{2}\ln(x^2 + 1)$, the square root being the power $\frac{1}{2}$. Differentiate term by term, the chain rule acting on the last two with inner functions $2x + 1$ and $x^2 + 1$: $f'(x) = \frac{3}{x} + \frac{4 \cdot 2}{2x + 1} - \frac{1}{2} \cdot \frac{2x}{x^2 + 1} = \frac{3}{x} + \frac{8}{2x + 1} - \frac{x}{x^2 + 1}$. At $x = 1$: $3 + \frac{8}{3} - \frac{1}{2} = \frac{18 + 16 - 3}{6} = \frac{31}{6}$. Differentiating the quotient directly works too, and takes half a page, with a product rule inside a quotient rule inside a chain rule.

b) On $(-1, 1)$, $1 + x > 0$ and $1 - x > 0$: $g(x) = \frac{1}{2}\left[\ln(1 + x) - \ln(1 - x)\right]$. Chain rule on the second term, inner function $1 - x$ with derivative $-1$: $g'(x) = \frac{1}{2}\left[\frac{1}{1 + x} + \frac{1}{1 - x}\right] = \frac{1}{2} \cdot \frac{(1 - x) + (1 + x)}{1 - x^2} = \frac{1}{1 - x^2}$. The sign of the second term is the trap: $-\ln(1 - x)$ differentiates to $-\frac{-1}{1 - x} = +\frac{1}{1 - x}$, and losing one of the two minus signs gives $\frac{x}{1 - x^2}$ instead.

c) $\ln(e^{3x}(x^2 + 1)) = \ln e^{3x} + \ln(x^2 + 1) = 3x + \ln(x^2 + 1)$, since $\ln$ and $\exp$ undo each other. Then $h'(x) = 3 + \frac{2x}{x^2 + 1}$. Without the simplification one needs the chain rule, the product rule and a fraction to simplify at the end: $\frac{3e^{3x}(x^2 + 1) + 2x e^{3x}}{e^{3x}(x^2 + 1)}$, which reduces to the same thing. Simplifying first is not an elegance, it removes the places where marks are lost.

d) No. $\ln(x^2)$ is defined wherever $x^2 > 0$, that is for all $x \ne 0$, while $2 \ln x$ needs $x > 0$. The correct law is $\ln(x^2) = 2\ln|x|$: the two functions agree for $x > 0$ (the blue branch) and only $\ln(x^2)$ has the orange branch on the left. Their derivatives: $\frac{d}{dx}\ln(x^2) = \frac{2x}{x^2} = \frac{2}{x}$ for all $x \ne 0$, and $\frac{d}{dx}2\ln x = \frac{2}{x}$ for $x > 0$. The same formula, on different domains: at $x = -1$, the first has derivative $-2$ and the second does not exist.

e) At $x = 0$ the fraction equals $\frac{1 \cdot 27}{32} > 0$, so $k$ is defined near $0$. In general the three factors do not have a fixed sign, so the law is written with absolute values: $k(x) = 2\ln|x + 1| + 3\ln|x + 3| - 5\ln|x + 2|$, valid wherever the fraction is positive. By $\frac{d}{dx}\ln|u| = \frac{u'}{u}$: $k'(x) = \frac{2}{x + 1} + \frac{3}{x + 3} - \frac{5}{x + 2}$, so $k'(0) = 2 + 1 - \frac{5}{2} = \frac{1}{2}$. Near $x = 0$ the absolute values are harmless, but at $x = -4$, where the fraction equals $\frac{9 \cdot (-1)}{-32} = \frac{9}{32} > 0$ and $k$ is defined, writing $\ln(x + 1)$ would be the logarithm of $-3$.

a) For $x > 0$, $y > 0$, and $\ln y = \frac{1}{2}\ln x + 2\ln(x + 3) - 3\ln(x + 1)$. Differentiate both sides with respect to $x$. On the left, $\ln y$ is a composition whose inner function is $y(x)$, so the chain rule gives $\frac{y'}{y}$; on the right, term by term: $\frac{y'}{y} = \frac{1}{2x} + \frac{2}{x + 3} - \frac{3}{x + 1}$. Each term is the relative rate of one factor, multiplied by its exponent: the square root contributes $\frac{1}{2}$, the denominator contributes with a minus sign.

b) $y(1) = \frac{1 \cdot 16}{8} = 2$. From a): $\frac{y'(1)}{y(1)} = \frac{1}{2} + \frac{2}{4} - \frac{3}{2} = -\frac{1}{2}$, so $y'(1) = 2 \cdot \left(-\frac{1}{2}\right) = -1$. Tangent: $y = 2 - (x - 1) = -x + 3$. The most common loss here is to stop at $\frac{y'}{y} = -\frac{1}{2}$ and call it the slope: the last step, multiplying by $y$, is part of the method, and forgetting it gives the relative rate instead of the derivative.

c) For $x < 2$, $(x - 2)^3 < 0$ while $(x + 1)^2 > 0$, so $y < 0$ on that whole region (except at $x = -1$, excluded): $\ln y$ does not exist there, in particular at $x = 0$ where $y = -8$. Use $\ln|y| = 3\ln|x - 2| - 2\ln|x + 1|$; differentiating, $\frac{y'}{y} = \frac{3}{x - 2} - \frac{2}{x + 1}$, valid wherever $y \ne 0$. At $x = 0$: $\frac{y'}{y} = -\frac{3}{2} - 2 = -\frac{7}{2}$, so $y'(0) = -8 \cdot \left(-\frac{7}{2}\right) = 28$. Quotient rule check: $y' = \frac{3(x - 2)^2 (x + 1)^2 - 2(x + 1)(x - 2)^3}{(x + 1)^4}$, and at $0$: $\frac{3 \cdot 4 \cdot 1 - 2 \cdot 1 \cdot (-8)}{1} = 12 + 16 = 28$. The two methods agree, as they must.

d) Put the bracket over a common denominator: $\frac{3}{x - 2} - \frac{2}{x + 1} = \frac{3(x + 1) - 2(x - 2)}{(x - 2)(x + 1)} = \frac{x + 7}{(x - 2)(x + 1)}$. Then $y' = \frac{(x - 2)^3}{(x + 1)^2} \cdot \frac{x + 7}{(x - 2)(x + 1)} = \frac{(x - 2)^2 (x + 7)}{(x + 1)^3}$. This simplified formula is valid for every $x \ne -1$, including $x = 2$. It vanishes at $x = 2$ and at $x = -7$: horizontal tangents at $(2, 0)$ and at $(-7, \frac{(-9)^3}{36}) = (-7, -\frac{81}{4})$. The formula $y' = y(\dots)$ was derived by dividing by $y$, which is illegal where $y = 0$: the point $x = 2$ is lost by the method and recovered only by simplifying. The figure shows the curve negative on the whole left part, and flat at both points.

e) Let $P = fg > 0$. Then $\ln P = \ln f + \ln g$, and differentiating, $\frac{P'}{P} = \frac{f'}{f} + \frac{g'}{g}$. Multiply by $P = fg$: $P' = f'g + fg'$, the product rule. With $Q = \frac{f}{g}$: $\ln Q = \ln f - \ln g$, so $\frac{Q'}{Q} = \frac{f'}{f} - \frac{g'}{g}$, and $Q' = \frac{f}{g}\left(\frac{f'}{f} - \frac{g'}{g}\right) = \frac{f'}{g} - \frac{fg'}{g^2} = \frac{f'g - fg'}{g^2}$, the quotient rule. In words: relative rates ADD under a product and SUBTRACT under a quotient. That sentence is the whole method.

a) For $x > 0$, $y > 0$, and $\ln y = x \ln x$. Differentiate both sides: on the left the chain rule gives $\frac{y'}{y}$; on the right the PRODUCT rule gives $1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$. So $y' = y(\ln x + 1) = x^x(\ln x + 1)$. At $x = 1$: $y'(1) = 1 \cdot (0 + 1) = 1$. The second student gets $1^1 \ln 1 = 0$: wrong. The first gets $1^1 = 1$, right by accident; test at $x = e$ instead, where the correct value is $e^e(1 + 1) = 2e^e$ and the first student's is $e^e$. Each student applied a rule outside its hypotheses: the first treated the exponent as a constant, the second treated the base as a constant. The correct answer is exactly the SUM of their two answers, one term per place where $x$ appears.

b) $y' = x^x(\ln x + 1)$ and $x^x > 0$, so $y' = 0$ exactly when $\ln x = -1$, that is $x = e^{-1} = \frac{1}{e}$. There $y = \left(\frac{1}{e}\right)^{1/e} = e^{-1/e}$: the horizontal tangent is at $\left(\frac{1}{e}, e^{-1/e}\right)$. At $x = 1$: $y(1) = 1$ and $y'(1) = 1$, so the tangent is $y = 1 + (x - 1) = x$. The figure shows both tangents: the curve dips to its low point at $x = \frac{1}{e}$, then crosses back up through $(1, 1)$ with slope $1$.

c) On $(0, \pi)$, $\sin x > 0$, so $\ln y = x \ln(\sin x)$ is defined. Product rule on the right, with the chain rule on $\ln(\sin x)$, inner function $\sin x$: $\frac{y'}{y} = \ln(\sin x) + x \cdot \frac{\cos x}{\sin x}$. Hence $y' = (\sin x)^x\left(\ln(\sin x) + x \cot x\right)$. At $x = \frac{\pi}{2}$: $\sin\frac{\pi}{2} = 1$, so $\ln 1 = 0$, and $\cot\frac{\pi}{2} = 0$: $y' = 1 \cdot (0 + 0) = 0$. The tangent is horizontal at $\left(\frac{\pi}{2}, 1\right)$. Outside $(0, \pi)$ the base is negative or zero and $(\sin x)^x$ is not even defined for most $x$: the interval is part of the question.

d) For $x > 0$: $\ln y = \ln x \cdot \ln x = (\ln x)^2$. Chain rule on the right, outer function the square, inner function $\ln x$: $\frac{y'}{y} = 2\ln x \cdot \frac{1}{x}$. So $y' = \frac{2 \ln x}{x}\,x^{\ln x}$. At $x = e$: $y(e) = e^{\ln e} = e^1 = e$ and $\frac{y'}{y} = \frac{2}{e}$, so $y'(e) = e \cdot \frac{2}{e} = 2$. Note that $(\ln x)^2$ is NOT $\ln(x^2) = 2\ln x$: confusing them here gives $\frac{y'}{y} = \frac{2}{x}$, and $y'(e) = 2$ by coincidence at $x = e$ but a wrong derivative almost everywhere else: at $x = e^2$ the true value is $\frac{4}{e^2} \cdot e^4 = 4e^2$ against $\frac{2}{e^2} \cdot e^4 = 2e^2$.

e) $x^e$: constant exponent $e$, power rule, $ex^{e-1}$. $e^x$: its own derivative, $e^x$. $e^e$: a CONSTANT, a number close to $15$, derivative $0$. $x^x$: variable base and exponent, logarithmic differentiation, $x^x(\ln x + 1)$. At $x = e$: $e \cdot e^{e-1} = e^e$, then $e^e$, then $0$, then $e^e \cdot 2 = 2e^e$. The first two coincide at $x = e$, a nice fact and nothing more; the third is the one most often got wrong, by students who apply the power rule to a number. Before differentiating any power, ask two questions: does the base contain $x$? Does the exponent contain $x$? Two yes answers mean: take the logarithm.