Exercise 1: The derivative of ln: the domain first, the inner function named

9 answers to check

Section 3.8 of Thomas gives $\frac{d}{dx}\ln x = \frac{1}{x}$ for $x > 0$ and, with the chain rule, $\frac{d}{dx}\ln u = \frac{u'}{u}$ wherever $u > 0$ . With an absolute value, $\frac{d}{dx}\ln|u| = \frac{u'}{u}$ wherever $u \ne 0$ : the formula keeps no absolute value.

Every question on a logarithm starts with its DOMAIN, read on the function and not on its derivative. The figure shows $y = \ln(4 - x^2)$ between its two dashed boundaries.

a) Give the domain of $f(x) = \ln(4 - x^2)$ , differentiate $f$ , compute $f'(1)$ and find where the tangent is horizontal.
b) Differentiate $g(x) = \ln|2x - 5|$ and compute $g'(2)$ . Why would $\ln(2x - 5)$ give no answer at $x = 2$ ?
c) Differentiate $h(x) = x\ln x - x$ and compute $h'(e^2)$ .
d) Give the domain of $k(x) = \sqrt{\ln x}$ , differentiate it, and compute $k'(e)$ to four decimals.
e) Thomas obtains $\frac{d}{dx}\ln x$ from the identity $e^{\ln x} = x$ . Redo this derivation, then write the tangent line to $y = \ln x$ at $x = e$ and show that it passes through the origin.

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Answers

a) Domain $(-2, 2)$ ; $f'(x) = -\frac{2x}{4 - x^2}$ , $f'(1) = -\frac{2}{3}$ ; horizontal tangent at $x = 0$
b) $g'(x) = \frac{2}{2x - 5}$ , $g'(2) = -2$ ; $\ln(2x - 5)$ is undefined for $x \le \frac{5}{2}$
c) $h'(x) = \ln x$ , $h'(e^2) = 2$
d) Domain $[1, \infty)$ ; $k'(x) = \frac{1}{2x\sqrt{\ln x}}$ for $x > 1$ , $k'(e) = \frac{1}{2e} \approx 0.1839$
e) $y = \ln x \iff e^y = x$ , so $e^y y' = 1$ and $y' = \frac{1}{x}$ ; tangent $y = \frac{x}{e}$ , intercept $0$

a)

The logarithm needs $4 - x^2 > 0$ , that is $x^2 < 4$ , that is $-2 < x < 2$ . The algebra slip that costs the domain mark is to write $x < \pm 2$ or $x < 2$ : a square smaller than $4$ traps $x$ between $-2$ and $2$ , which is what the dashed lines show. Chain rule with inner function $u = 4 - x^2$ , $u' = -2x$ : $f'(x) = \frac{u'}{u} = -\frac{2x}{4 - x^2}$ . At $x = 1$ : $f'(1) = -\frac{2}{3}$ . The denominator is positive on the whole domain, so $f'(x) = 0$ exactly when $x = 0$ : horizontal tangent at $(0, \ln 4)$ , the top of the arch on the figure. Writing $\frac{1}{4 - x^2}$ , the derivative of $\ln$ without the derivative of the inside, is the most frequent loss.

b)

$g$ is defined wherever $2x - 5 \ne 0$ , that is $x \ne \frac{5}{2}$ . By $\frac{d}{dx}\ln|u| = \frac{u'}{u}$ with $u = 2x - 5$ , $u' = 2$ : $g'(x) = \frac{2}{2x - 5}$ , with NO absolute value in the answer. At $x = 2$ : $g'(2) = \frac{2}{-1} = -2$ . The student who writes $\frac{2}{|2x - 5|}$ gets $+2$ , the wrong sign: on the left branch $|2x - 5| = 5 - 2x$ decreases as $x$ increases, and so does its logarithm. Without the absolute value, $\ln(2x - 5)$ exists only for $x > \frac{5}{2}$ , and $g'(2)$ would be a number about a point where there is no function.

c)

Product rule on $x\ln x$ : $(x)'\ln x + x(\ln x)' = \ln x + x \cdot \frac{1}{x} = \ln x + 1$ . Subtract the derivative of $x$ : $h'(x) = \ln x + 1 - 1 = \ln x$ , for $x > 0$ . At $x = e^2$ : $h'(e^2) = \ln e^2 = 2$ , since $\ln$ and $\exp$ undo each other. The simplification $x \cdot \frac{1}{x} = 1$ looks trivial; forgetting it leaves an answer the marker has to decode.

d)

The square root needs $\ln x \ge 0$ , that is $x \ge 1$ , and $\ln$ itself needs $x > 0$ : domain $[1, \infty)$ . Chain rule, outer function the square root, inner function $u = \ln x$ : $k'(x) = \frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}$ , valid for $x > 1$ only: at $x = 1$ the denominator is $0$ , and the graph leaves $(1, 0)$ with a vertical tangent. At $x = e$ : $\ln e = 1$ , so $k'(e) = \frac{1}{2e} \approx 0.1839$ . Rounding comes LAST: the exact value $\frac{1}{2e}$ is the answer, the decimal is what the calculator reports.

e)

Let $y = \ln x$ for $x > 0$ . By definition of $\ln$ as the inverse of $e^x$ , $e^y = x$ . Differentiate both sides with respect to $x$ , $y$ being a function of $x$ (implicit differentiation, section 3.7): $e^y \cdot y' = 1$ . Since $e^y > 0$ , divide: $y' = \frac{1}{e^y} = \frac{1}{x}$ , the last step replacing $e^y$ by $x$ . At $x = e$ : the point is $(e, \ln e) = (e, 1)$ and the slope is $\frac{1}{e}$ . Tangent: $y = 1 + \frac{1}{e}(x - e) = 1 + \frac{x}{e} - 1 = \frac{x}{e}$ . Its intercept is $0$ : it passes through the origin. The algebra step $\frac{1}{e} \cdot e = 1$ is the whole proof; a student who leaves $y - 1 = \frac{1}{e}(x - e)$ unsimplified has the right line and cannot see the origin in it.

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Exercise 2: The laws of logarithms BEFORE differentiating, and only the true ones

9 answers to check

For $a > 0$ and $b > 0$ : $\ln(ab) = \ln a + \ln b$ , $\ln\frac{a}{b} = \ln a - \ln b$ , $\ln(a^r) = r\ln a$ , and $\ln e^u = u$ . A root is a fractional power, $\sqrt[3]{x} = x^{1/3}$ ; a factor in the denominator comes out with a MINUS sign. Expanding first turns one quotient rule inside a chain rule into a sum of short terms.

Nothing splits $\ln(a + b)$ . And the laws need POSITIVE arguments: with factors of unknown sign they are written with absolute values.

a) For $x > 0$ , differentiate $f(x) = \ln\frac{\sqrt[3]{x}\,(x^2 + 3)^2}{(5x + 1)^4}$ and compute $f'(1)$ .
b) Differentiate $g(x) = \ln\frac{e^{2x}}{1 + e^{2x}}$ , simplify to a single fraction, and compute $g'(0)$ .
c) For $x > 0$ , differentiate $h(x) = \log_2\frac{8x^3}{\sqrt{x + 1}}$ and compute $h'(1)$ to four decimals.
d) Say whether each rewriting is legitimate:(i) $\ln(x^2 + 9) = 2\ln x + \ln 9$ (ii) $\ln\sqrt{x^2 + 9} = \frac{1}{2}\ln(x^2 + 9)$ (iii) $\ln(9x^2) = \ln 9 + 2\ln|x|$ for $x \ne 0$ .Then compute the true derivative of $\ln(x^2 + 9)$ at $x = 3$ and of $\ln(9x^2)$ at $x = -3$ .
e) Let $F(x) = \ln\left((x - 1)(x - 4)\right)$ . Give its domain, explain why $F(x) = \ln(x - 1) + \ln(x - 4)$ is wrong at $x = 0$ , write the correct expansion, and compute $F'(0)$ .

Type your answers, the page tells you right or wrong 0/9

a)

f'(1)

b)

g'(0)

c)

h'(1)

d)

Rewriting (i)

Rewriting (ii)

Rewriting (iii)

Derivative of

\ln(x^2 + 9)

at

x = 3

Derivative of

\ln(9x^2)

at

x = -3

e)

F'(0)

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Answers

a) $f'(x) = \frac{1}{3x} + \frac{4x}{x^2 + 3} - \frac{20}{5x + 1}$ , $f'(1) = -2$
b) $g(x) = 2x - \ln(1 + e^{2x})$ , $g'(x) = \frac{2}{1 + e^{2x}}$ , $g'(0) = 1$
c) $h(x) = 3 + 3\log_2 x - \frac{1}{2}\log_2(x + 1)$ , $h'(1) = \frac{11}{4\ln 2} \approx 3.9674$
d) (i) not legitimate, (ii) and (iii) legitimate; derivatives $\frac{1}{3}$ at $x = 3$ and $-\frac{2}{3}$ at $x = -3$
e) Domain $x < 1$ or $x > 4$ ; $F(x) = \ln|x - 1| + \ln|x - 4|$ , $F'(0) = -\frac{5}{4}$

a)

For $x > 0$ every factor is positive, so the laws apply as they are. The cube root is the power $\frac{1}{3}$ and the denominator comes out with a minus sign: $f(x) = \frac{1}{3}\ln x + 2\ln(x^2 + 3) - 4\ln(5x + 1)$ . This line is the algebra where the marks are: a student who writes $3\ln x$ for the cube root, or $+4\ln(5x + 1)$ , has lost the question before differentiating anything. Term by term, chain rule on the last two with inner functions $x^2 + 3$ and $5x + 1$ : $f'(x) = \frac{1}{3x} + \frac{2 \cdot 2x}{x^2 + 3} - \frac{4 \cdot 5}{5x + 1} = \frac{1}{3x} + \frac{4x}{x^2 + 3} - \frac{20}{5x + 1}$ . At $x = 1$ : $\frac{1}{3} + 1 - \frac{20}{6} = \frac{1}{3} + 1 - \frac{10}{3} = -2$ .

b)

$\ln\frac{e^{2x}}{1 + e^{2x}} = \ln e^{2x} - \ln(1 + e^{2x}) = 2x - \ln(1 + e^{2x})$ , since $\ln e^{u} = u$ ; both arguments are positive for every $x$ . Chain rule on the second term, inner function $1 + e^{2x}$ with derivative $2e^{2x}$ : $g'(x) = 2 - \frac{2e^{2x}}{1 + e^{2x}}$ . Common denominator: $g'(x) = \frac{2(1 + e^{2x}) - 2e^{2x}}{1 + e^{2x}} = \frac{2}{1 + e^{2x}}$ . At $x = 0$ : $g'(0) = \frac{2}{2} = 1$ . The fake law $\ln(1 + e^{2x}) = \ln 1 + \ln e^{2x} = 2x$ would make $g$ identically $0$ , a function whose graph is the $x$ axis: a quick sanity check kills it, since $g(0) = \ln\frac{1}{2} \ne 0$ .

c)

For $x > 0$ : $h(x) = \log_2 8 + 3\log_2 x - \frac{1}{2}\log_2(x + 1) = 3 + 3\log_2 x - \frac{1}{2}\log_2(x + 1)$ , because $\log_2 8 = 3$ is a CONSTANT. Change of base, $\log_2 u = \frac{\ln u}{\ln 2}$ , so every term carries $\frac{1}{\ln 2}$ : $h'(x) = \frac{1}{\ln 2}\left(\frac{3}{x} - \frac{1}{2(x + 1)}\right)$ . At $x = 1$ : $\frac{1}{\ln 2}\left(3 - \frac{1}{4}\right) = \frac{11}{4\ln 2} \approx 3.9674$ . Two frequent losses: differentiating $\log_2 8$ as if it were a function, and dropping the $\frac{1}{\ln 2}$ , which gives $2.75$ .

d)

(i) NOT legitimate: $x^2 + 9$ is a SUM, and no law splits the logarithm of a sum; $2\ln x + \ln 9 = \ln(9x^2)$ , a different function. The true derivative, chain rule with inner function $x^2 + 9$ : $\frac{2x}{x^2 + 9}$ , at $x = 3$ : $\frac{6}{18} = \frac{1}{3}$ , while the fake version gives $\frac{2}{x} = \frac{2}{3}$ . (ii) Legitimate: the power law with exponent $\frac{1}{2}$ , and $x^2 + 9 > 0$ for every $x$ . (iii) Legitimate: $9x^2 = 9|x|^2$ with $9 > 0$ and $|x| > 0$ , so $\ln(9x^2) = \ln 9 + 2\ln|x|$ for every $x \ne 0$ ; writing $2\ln x$ would delete the whole left half of the domain. Its derivative is $\frac{2}{x}$ , at $x = -3$ : $-\frac{2}{3}$ .

e)

The logarithm needs $(x - 1)(x - 4) > 0$ : both factors of the same sign, so $x < 1$ or $x > 4$ . At $x = 0$ the product is $(-1)(-4) = 4 > 0$ and $F(0) = \ln 4$ exists, but $\ln(0 - 1)$ and $\ln(0 - 4)$ do not: the law $\ln(ab) = \ln a + \ln b$ needs $a > 0$ and $b > 0$ , which holds only for $x > 4$ . On the whole domain: $F(x) = \ln|x - 1| + \ln|x - 4|$ . Then $F'(x) = \frac{1}{x - 1} + \frac{1}{x - 4}$ and $F'(0) = -1 - \frac{1}{4} = -\frac{5}{4}$ . Check without expanding: $F(x) = \ln(x^2 - 5x + 4)$ , $F'(x) = \frac{2x - 5}{x^2 - 5x + 4}$ , and at $0$ : $-\frac{5}{4}$ . The two routes agree, as they must.

Exercise 3: The derivative of an inverse function at a point, without a formula for the inverse

8 answers to check

Thomas 3.8: if $f$ is differentiable on an interval and $f'$ is never zero there, then $f^{-1}$ is differentiable and, at $b = f(a)$ , $(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}$ . The graphs of $f$ and $f^{-1}$ are mirror images in the line $y = x$ : the point $(a, b)$ of $f$ becomes the point $(b, a)$ of $f^{-1}$ , and the slope is inverted.

Let $f(x) = e^x + 2x$ . The equation $e^x + 2x = y$ cannot be solved for $x$ with algebra, so no formula for $f^{-1}$ exists on the page. The figure shows $f$ , $f^{-1}$ and the line $y = x$ .

a) Explain why $f$ is one-to-one, with no derivative test, and find $f^{-1}(1)$ .
b) Compute $(f^{-1})'(1)$ , then the tangent line to $y = f^{-1}(x)$ at the point $(1, 0)$ .
c) A student answers $(f^{-1})'(1) = \frac{1}{f'(1)} = \frac{1}{e + 2}$ . What did he do, and which point is his number actually about?
d) Let $q(x) = x^2 - 6x$ restricted to $x \ge 3$ . Compute $(q^{-1})'(16)$ with the theorem, then find $q^{-1}$ explicitly by completing the square and check. What happens at $b = -9$ ?
e) For $p(x) = x^4$ on $x > 0$ , $p^{-1}(x) = x^{1/4}$ . Compute $(p^{-1})'(81)$ with the theorem, then with the power rule.

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Answers

a) $e^x$ and $2x$ are strictly increasing, so is their sum: one-to-one; $f(0) = 1$ , so $f^{-1}(1) = 0$
b) $(f^{-1})'(1) = \frac{1}{f'(0)} = \frac{1}{3}$ ; tangent $y = \frac{1}{3}x - \frac{1}{3}$
c) He evaluated $f'$ at $b = 1$ instead of $a = 0$ ; $\frac{1}{e + 2}$ is the slope of $f^{-1}$ at $b = e + 2$
d) $q(8) = 16$ , $(q^{-1})'(16) = \frac{1}{q'(8)} = \frac{1}{10}$ ; $q^{-1}(x) = 3 + \sqrt{x + 9}$ ; at $b = -9$ , $q'(3) = 0$ : vertical tangent
e) $(p^{-1})'(81) = \frac{1}{p'(3)} = \frac{1}{108}$ , and $\frac{1}{4} \cdot 81^{-3/4} = \frac{1}{108}$

a)

If $x_1 < x_2$ , then $e^{x_1} < e^{x_2}$ and $2x_1 < 2x_2$ ; adding, $f(x_1) < f(x_2)$ . So $f$ is strictly increasing, and a horizontal line meets its graph at most once: $f$ is one-to-one (horizontal line test, Thomas 1.6). No derivative is needed, and none is allowed for this step at this point of the course. Since $f(0) = e^0 + 0 = 1$ , the point $(0, 1)$ is on $f$ and $f^{-1}(1) = 0$ . Finding $a$ by inspection is the normal way: try $x = 0$ , $x = 1$ , $x = -1$ before anything else.

b)

$f'(x) = e^x + 2$ , never zero, so the theorem applies. With $b = 1$ and $a = f^{-1}(1) = 0$ : $f'(0) = 1 + 2 = 3$ , and $(f^{-1})'(1) = \frac{1}{f'(0)} = \frac{1}{3}$ . The tangent to $f^{-1}$ at $(1, 0)$ : $y = 0 + \frac{1}{3}(x - 1) = \frac{1}{3}x - \frac{1}{3}$ . On the solution figure, the tangent to $f$ at $(0, 1)$ has slope $3$ and its mirror image, the tangent to $f^{-1}$ at $(1, 0)$ , has slope $\frac{1}{3}$ : reflecting a line in $y = x$ swaps rise and run.

c)

He evaluated $f'$ at $b = 1$ , the input of $f^{-1}$ , instead of at $a = f^{-1}(1) = 0$ , the input of $f$ . The theorem pairs the two graphs point by point: the slope of $f^{-1}$ at $(b, a)$ is the reciprocal of the slope of $f$ at the SWAPPED point $(a, b)$ . His number $\frac{1}{f'(1)} = \frac{1}{e + 2} \approx 0.2119$ is a true slope of $f^{-1}$ , but at $b = f(1) = e + 2$ , a different point. Writing the pair $(a, b)$ on the copy before any derivative is the gesture that prevents this loss, which costs the whole question.

d)

$q(x) = (x - 3)^2 - 9$ is a parabola with vertex at $x = 3$ , so on $x \ge 3$ it is one-to-one (restricting the domain, Thomas 1.6). Solve $q(a) = 16$ : $a^2 - 6a - 16 = 0$ , $(a - 8)(a + 2) = 0$ , and $a = -2$ is rejected since $a \ge 3$ : $a = 8$ . $q'(x) = 2x - 6$ , $q'(8) = 10$ , so $(q^{-1})'(16) = \frac{1}{10}$ . Explicitly: $y = x^2 - 6x$ gives $y + 9 = (x - 3)^2$ , so $x = 3 + \sqrt{y + 9}$ , with the PLUS sign because $x \ge 3$ ; hence $q^{-1}(x) = 3 + \sqrt{x + 9}$ . Its derivative is $\frac{1}{2\sqrt{x + 9}}$ , and at $16$ : $\frac{1}{2 \cdot 5} = \frac{1}{10}$ . The theorem and the explicit route agree. At $b = -9 = q(3)$ , $q'(3) = 0$ : the hypothesis fails, and indeed $\frac{1}{2\sqrt{x + 9}}$ blows up at $x = -9$ , where $q^{-1}$ has a VERTICAL tangent, the mirror image of the horizontal tangent of $q$ at its vertex.

e)

$p^{-1}(81) = 3$ , since $3^4 = 81$ and $3 > 0$ . $p'(x) = 4x^3$ , $p'(3) = 108$ , so $(p^{-1})'(81) = \frac{1}{108}$ . Power rule: $\frac{d}{dx}x^{1/4} = \frac{1}{4}x^{-3/4}$ , and $81^{-3/4} = \left(81^{1/4}\right)^{-3} = 3^{-3} = \frac{1}{27}$ , so the value is $\frac{1}{4} \cdot \frac{1}{27} = \frac{1}{108}$ . The rewriting of $81^{-3/4}$ , root first and power second, is where a MATH 203 copy loses the mark: taking $81^3$ first gives a number no one roots by hand, and the negative exponent must end up in the denominator.

Exercise 4: Logarithms in other bases: the constant ln a, even when the base moves

9 answers to check

For $a > 0$ , $a \ne 1$ : $\log_a x = \frac{\ln x}{\ln a}$ , so $\frac{d}{dx}\log_a u = \frac{u'}{u\ln a}$ . The change of base is the whole method: it turns every logarithm into $\ln$ divided by a constant, or, when the base itself contains $x$ , into a QUOTIENT of two natural logarithms. The base never disappears: it survives as $\ln a$ , positive for $a > 1$ , negative for $0 < a < 1$ ; and $\frac{d}{dx}a^x = a^x\ln a$ (section 3.6) shows the same constant as a slope.

The figure shows $y = 2^x$ , $y = e^x$ and $y = 4^x$ , all through $(0, 1)$ , and the dashed line $y = 1 + x$ .

a) Differentiate $k(x) = \log_4(x^2 + 12)$ and compute $k'(2)$ to four decimals.
b) For $x > 0$ , differentiate $g(x) = x^2\log_3 x$ , factor the result, and compute $g'(3)$ to four decimals.
c) For $x > 1$ , rewrite $h(x) = \log_x e$ with natural logarithms, differentiate it, and compute $h'(e)$ to four decimals.
d) The dashed line $y = 1 + x$ is tangent to one of the three curves at $(0, 1)$ . Which one? Give the slopes of the other two at $x = 0$ , and the point where the tangent to $y = 4^x$ at $x = 0$ cuts the $x$ axis.
e) Simplify, then differentiate: $m(x) = \log_2 x \cdot \log_x 8$ for $x > 0$ , $x \ne 1$ , and $n(x) = \log_3(9^x)$ .

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Answers

a) $k'(x) = \frac{2x}{(x^2 + 12)\ln 4}$ , $k'(2) = \frac{1}{4\ln 4} = \frac{1}{8\ln 2} \approx 0.1803$
b) $g'(x) = x\left(2\log_3 x + \frac{1}{\ln 3}\right)$ , $g'(3) = 6 + \frac{3}{\ln 3} \approx 8.7307$
c) $h(x) = \frac{1}{\ln x}$ , $h'(x) = -\frac{1}{x(\ln x)^2}$ , $h'(e) = -\frac{1}{e} \approx -0.3679$
d) $y = e^x$ ; slopes $\ln 2 \approx 0.6931$ and $\ln 4 \approx 1.3863$ ; the tangent $y = 1 + x\ln 4$ cuts the axis at $x = -\frac{1}{\ln 4} \approx -0.7213$
e) $m(x) = 3$ , $m'(x) = 0$ ; $n(x) = 2x$ , $n'(x) = 2$

a)

Change of base: $k(x) = \frac{\ln(x^2 + 12)}{\ln 4}$ , and $\frac{1}{\ln 4}$ is a constant factor. Chain rule with inner function $x^2 + 12$ : $k'(x) = \frac{1}{\ln 4} \cdot \frac{2x}{x^2 + 12}$ . At $x = 2$ : $\frac{4}{16\ln 4} = \frac{1}{4\ln 4}$ , and since $\ln 4 = \ln 2^2 = 2\ln 2$ , this is $\frac{1}{8\ln 2} \approx 0.1803$ . The domain is all of $\mathbb{R}$ , since $x^2 + 12 > 0$ . Writing $\frac{2x}{4(x^2 + 12)}$ , the base in place of its logarithm, gives $0.25$ and costs the mark.

b)

Product rule, with $(\log_3 x)' = \frac{1}{x\ln 3}$ : $g'(x) = 2x\log_3 x + x^2 \cdot \frac{1}{x\ln 3} = 2x\log_3 x + \frac{x}{\ln 3}$ , the simplification $\frac{x^2}{x} = x$ being the step to write. Factor $x$ : $g'(x) = x\left(2\log_3 x + \frac{1}{\ln 3}\right)$ . At $x = 3$ : $\log_3 3 = 1$ , so $g'(3) = 3\left(2 + \frac{1}{\ln 3}\right) = 6 + \frac{3}{\ln 3} \approx 8.7307$ . The two frequent losses: $\frac{1}{3x}$ for the derivative of $\log_3 x$ , which gives $7$ , and $\log_3 3$ left unevaluated, as if it were a mystery number.

c)

Here the BASE is the variable, so the rule $\frac{1}{x\ln a}$ does not apply: it needs a constant base. Change of base: $h(x) = \log_x e = \frac{\ln e}{\ln x} = \frac{1}{\ln x}$ , defined for $x > 0$ , $x \ne 1$ , and here $x > 1$ . Rewrite as a power, $h = (\ln x)^{-1}$ , and use the chain rule with inner function $\ln x$ : $h'(x) = -(\ln x)^{-2} \cdot \frac{1}{x} = -\frac{1}{x(\ln x)^2}$ . At $x = e$ : $-\frac{1}{e \cdot 1} = -\frac{1}{e} \approx -0.3679$ . The copy that answers $\frac{1}{e\ln x}$ , treating $x$ as the constant and $e$ as the variable, has read the rule backwards: before any rule, ask which of base and argument contains $x$ .

d)

The slope of $a^x$ at $x = 0$ is $a^0\ln a = \ln a$ . It equals $1$ exactly when $a = e$ : the line $y = 1 + x$ , of slope $1$ through $(0, 1)$ , is the tangent to $y = e^x$ . For $2^x$ the slope at $0$ is $\ln 2 \approx 0.6931 < 1$ , flatter than the dashed line; for $4^x$ it is $\ln 4 \approx 1.3863 > 1$ , steeper, as the figure shows. The tangent to $4^x$ at $0$ is $y = 1 + x\ln 4$ ; it cuts the $x$ axis where $1 + x\ln 4 = 0$ , at $x = -\frac{1}{\ln 4} = -\frac{1}{2\ln 2} \approx -0.7213$ , to the right of $-1$ , where the dashed line cuts it: a steeper tangent reaches the axis sooner. This is how Thomas characterizes $e$ : the one base whose exponential crosses the $y$ axis with slope exactly $1$ , which is why $\frac{d}{dx}e^x = e^x$ with no constant.

e)

By change of base, $m(x) = \frac{\ln x}{\ln 2} \cdot \frac{\ln 8}{\ln x} = \frac{\ln 8}{\ln 2} = \frac{3\ln 2}{\ln 2} = 3$ for $x > 0$ , $x \ne 1$ : $m$ is constant and $m'(x) = 0$ on its domain. And $n(x) = \log_3(9^x) = x\log_3 9 = 2x$ by the power law, so $n'(x) = 2$ . The product rule on $m$ and the chain rule on $n$ lead to the same answers after half a page of fractions; rewriting first is not elegance, it removes the places where marks are lost.

Exercise 5: Logarithmic differentiation of heavy products and quotients

10 answers to check

Logarithmic differentiation, Thomas 3.8: take $\ln|y|$ of both sides, expand with the laws of logarithms, differentiate both sides (the left side gives $\frac{y'}{y}$ by the chain rule), then multiply by $y$ . The absolute value makes the method valid wherever $y \ne 0$ .

The expansion is the algebra step where the marks are: each exponent in front, roots as fractional powers, the denominator with minus signs, and $\ln e^{u} = u$ .

a) Let $y = \frac{(x + 1)^4\sqrt{2x + 7}}{(x^2 + 1)^3}$ for $x > -1$ . Find $\frac{y'}{y}$ , then $y(1)$ , $y'(1)$ and the tangent line at $x = 1$ .
b) Let $y = \frac{(x - 5)^3}{(x + 3)^2(x - 1)}$ . Explain why $\ln y$ cannot be used at $x = 3$ , and compute $y'(3)$ by logarithmic differentiation.
c) Let $y = \frac{(2x + 1)^5}{\sqrt[3]{x^2 + 7}\;e^{x}}$ for $x > -\frac{1}{2}$ . Compute $\frac{y'(1)}{y(1)}$ exactly, then $y'(1)$ to two decimals.
d) A student applies the method to $y = (x + 1)^4 + \sqrt{2x + 7}$ and writes $\ln y = 4\ln(x + 1) + \frac{1}{2}\ln(2x + 7)$ . Is the method needed here? Find the true $y'(1)$ .
e) For $x > 2$ , let $y = \frac{x^3}{(x - 2)^2}$ . By logarithmic differentiation, write $y'$ as a single fraction and find the point where the tangent is horizontal.

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Answers

a) $\frac{y'}{y} = \frac{4}{x + 1} + \frac{1}{2x + 7} - \frac{6x}{x^2 + 1}$ ; $y(1) = 6$ , $y'(1) = -\frac{16}{3}$ ; tangent $y = 6 - \frac{16}{3}(x - 1)$
b) $y(3) = -\frac{1}{9} < 0$ ; with $\ln|y|$ , $\frac{y'}{y} = -\frac{7}{3}$ at $3$ , so $y'(3) = \frac{7}{27}$
c) $\frac{y'(1)}{y(1)} = \frac{9}{4}$ , $y(1) = \frac{243}{2e}$ , $y'(1) = \frac{2187}{8e} \approx 100.57$
d) Not needed (a SUM: no law applies); $y' = 4(x + 1)^3 + \frac{1}{\sqrt{2x + 7}}$ , $y'(1) = \frac{97}{3}$
e) $y' = \frac{x^2(x - 6)}{(x - 2)^3}$ ; horizontal tangent at $\left(6, \frac{27}{2}\right)$

a)

For $x > -1$ every factor is positive, so $\ln y = 4\ln(x + 1) + \frac{1}{2}\ln(2x + 7) - 3\ln(x^2 + 1)$ . Differentiate both sides, chain rule on each term: $\frac{y'}{y} = \frac{4}{x + 1} + \frac{1}{2} \cdot \frac{2}{2x + 7} - \frac{3 \cdot 2x}{x^2 + 1} = \frac{4}{x + 1} + \frac{1}{2x + 7} - \frac{6x}{x^2 + 1}$ . At $x = 1$ : $y(1) = \frac{16 \cdot 3}{8} = 6$ and $\frac{y'}{y} = 2 + \frac{1}{9} - 3 = -\frac{8}{9}$ , so $y'(1) = 6 \cdot \left(-\frac{8}{9}\right) = -\frac{16}{3}$ . Tangent: $y = 6 - \frac{16}{3}(x - 1)$ . Stopping at $-\frac{8}{9}$ and calling it the slope forgets the last step: the method ends by MULTIPLYING BY $y$ .

b)

At $x = 3$ : $y(3) = \frac{(-2)^3}{6^2 \cdot 2} = -\frac{8}{72} = -\frac{1}{9} < 0$ , so $\ln y$ does not exist there, and $\ln(x - 5) = \ln(-2)$ neither. With absolute values, valid wherever $y \ne 0$ : $\ln|y| = 3\ln|x - 5| - 2\ln|x + 3| - \ln|x - 1|$ , so $\frac{y'}{y} = \frac{3}{x - 5} - \frac{2}{x + 3} - \frac{1}{x - 1}$ . At $3$ : $-\frac{3}{2} - \frac{2}{6} - \frac{1}{2} = -\frac{9 + 2 + 3}{6} = -\frac{7}{3}$ . Then $y'(3) = -\frac{1}{9} \cdot \left(-\frac{7}{3}\right) = \frac{7}{27}$ . Two negative signs meet in the last product: a sign lost in $y(3)$ gives $-\frac{7}{27}$ .

c)

All factors are positive for $x > -\frac{1}{2}$ . Expand, with the cube root as the power $\frac{1}{3}$ and $\ln e^x = x$ : $\ln y = 5\ln(2x + 1) - \frac{1}{3}\ln(x^2 + 7) - x$ . Differentiate: $\frac{y'}{y} = \frac{10}{2x + 1} - \frac{2x}{3(x^2 + 7)} - 1$ . At $x = 1$ : $\frac{10}{3} - \frac{2}{24} - 1 = \frac{40 - 1 - 12}{12} = \frac{27}{12} = \frac{9}{4}$ . And $y(1) = \frac{3^5}{\sqrt[3]{8}\,e} = \frac{243}{2e}$ , so $y'(1) = \frac{243}{2e} \cdot \frac{9}{4} = \frac{2187}{8e} \approx 100.57$ . The quotient rule here needs a product rule and two chain rules inside it; the logarithm reduces it to three one-line terms.

d)

Not needed, and not even legal as written. $y$ is a SUM, and no law splits the logarithm of a sum: $\ln\left((x + 1)^4 + \sqrt{2x + 7}\right)$ is not $4\ln(x + 1) + \frac{1}{2}\ln(2x + 7)$ , which is the logarithm of the PRODUCT. Differentiate term by term: $y' = 4(x + 1)^3 + \frac{1}{\sqrt{2x + 7}}$ , so $y'(1) = 32 + \frac{1}{3} = \frac{97}{3}$ . The student's line leads to $y' = y\left(2 + \frac{1}{9}\right) = 19 \cdot \frac{19}{9} = \frac{361}{9}$ at $x = 1$ , a wrong answer reached by a correct-looking method. Logarithmic differentiation is for products, quotients and powers only.

e)

For $x > 2$ , $y > 0$ and $\ln y = 3\ln x - 2\ln(x - 2)$ . Then $\frac{y'}{y} = \frac{3}{x} - \frac{2}{x - 2}$ . Common denominator, the algebra step that makes the zero visible: $\frac{3(x - 2) - 2x}{x(x - 2)} = \frac{x - 6}{x(x - 2)}$ . Multiply by $y$ : $y' = \frac{x^3}{(x - 2)^2} \cdot \frac{x - 6}{x(x - 2)} = \frac{x^2(x - 6)}{(x - 2)^3}$ . On $x > 2$ the factors $x^2$ and $(x - 2)^3$ are positive, so $y' = 0$ exactly at $x = 6$ , where $y = \frac{216}{16} = \frac{27}{2}$ . Horizontal tangent at $\left(6, \frac{27}{2}\right)$ . Whether this is a minimum is a question for chapter 4.

Exercise 6: A variable in the base AND in the exponent of the same power

9 answers to check

The power rule needs a constant exponent, the rule for $a^x$ a constant base. When $x$ sits in both places of ONE power $u^v$ , neither applies: take $\ln y = v\ln u$ , which turns the power into a product, differentiate with the product rule, and multiply by $y$ . The same computation can be read as $u^v = e^{v\ln u}$ .

The figure shows $y = x^{1/x}$ on $(0, 10]$ : it seems to have a high point somewhere near $x = 3$ .

a) For $x > 0$ , differentiate $y = x^{1/x}$ . Find exactly the point where the tangent is horizontal, and the tangent line at $x = 1$ .
b) Differentiate $y = (x^2 + 1)^x$ and compute $y'(1)$ to four decimals.
c) Differentiate $y = x^{x^2}$ for $x > 0$ and compute $y'(2)$ exactly. A student writes $y' = x^2 \cdot x^{x^2 - 1}$ : what does he get at $x = 2$ , and which term is missing?
d) For $x > 1$ , differentiate $y = (\ln x)^x$ and compute $y'(e)$ .
e) Does $y = 3^x x^3$ need logarithmic differentiation? Compute $y'(1)$ to four decimals.

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Answers

a) $y' = x^{1/x}\,\frac{1 - \ln x}{x^2}$ ; horizontal tangent at $\left(e, e^{1/e}\right) \approx (2.7183, 1.4447)$ ; tangent at $x = 1$ : $y = x$
b) $y' = (x^2 + 1)^x\left(\ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}\right)$ , $y'(1) = 2\ln 2 + 2 \approx 3.3863$
c) $y' = x^{x^2}(2x\ln x + x)$ , $y'(2) = 64\ln 2 + 32 \approx 76.3614$ ; the student gets $32$ , missing $64\ln 2$
d) $y' = (\ln x)^x\left(\ln(\ln x) + \frac{1}{\ln x}\right)$ , $y'(e) = 1$
e) No: $x$ is in the exponent of one factor and the base of another, so the product rule suffices; $y'(1) = 9 + 3\ln 3 \approx 12.2958$

a)

For $x > 0$ , $y > 0$ and $\ln y = \frac{1}{x}\ln x = \frac{\ln x}{x}$ . Quotient rule on the right: $\frac{y'}{y} = \frac{\frac{1}{x} \cdot x - \ln x \cdot 1}{x^2} = \frac{1 - \ln x}{x^2}$ , the simplification $\frac{1}{x} \cdot x = 1$ being the step to write. So $y' = x^{1/x}\,\frac{1 - \ln x}{x^2}$ . Since $x^{1/x} > 0$ and $x^2 > 0$ , $y' = 0$ exactly when $\ln x = 1$ , that is $x = e$ : the horizontal tangent is at $\left(e, e^{1/e}\right) \approx (2.7183, 1.4447)$ . The figure suggested $3$ ; the computation says $e$ . At $x = 1$ : $y = 1$ and $y' = 1 \cdot \frac{1 - 0}{1} = 1$ , so the tangent is $y = 1 + (x - 1) = x$ .

b)

$(x^2 + 1)^x > 0$ for every $x$ , so $\ln y = x\ln(x^2 + 1)$ . Product rule, with the chain rule on $\ln(x^2 + 1)$ , inner function $x^2 + 1$ : $\frac{y'}{y} = \ln(x^2 + 1) + x \cdot \frac{2x}{x^2 + 1}$ . So $y' = (x^2 + 1)^x\left(\ln(x^2 + 1) + \frac{2x^2}{x^2 + 1}\right)$ . At $x = 1$ : $y(1) = 2$ and $\frac{y'}{y} = \ln 2 + 1$ , so $y'(1) = 2\ln 2 + 2 \approx 3.3863$ . Reading it as $e^{x\ln(x^2 + 1)}$ and applying the chain rule gives the same line: the two methods are one.

c)

$\ln y = x^2\ln x$ . Product rule: $\frac{y'}{y} = 2x\ln x + x^2 \cdot \frac{1}{x}$ , and the algebra step is $x^2 \cdot \frac{1}{x} = x$ , a cancellation to write out on the copy rather than carry as $\frac{x^2}{x}$ . So $y' = x^{x^2}(2x\ln x + x)$ . At $x = 2$ : $y = 2^{4} = 16$ , the exponent $x^2 = 4$ computed FIRST, and $\frac{y'}{y} = 4\ln 2 + 2$ , so $y'(2) = 64\ln 2 + 32 \approx 76.3614$ . The student treats $x^2$ as a constant exponent: $4 \cdot 2^{3} = 32$ . He gets exactly the term $16 \cdot 2 = 32$ that comes from the BASE, and misses $16 \cdot 4\ln 2 = 64\ln 2$ , the term that comes from the EXPONENT. Reading $x^{x^2}$ as $(x^x)^2 = x^{2x}$ is the other algebra slip: at $x = 2$ it gives $16$ by coincidence, and at $x = 3$ it gives $729$ instead of $3^9 = 19\,683$ .

d)

The base $\ln x$ must be positive, hence $x > 1$ . $\ln y = x\ln(\ln x)$ . Product rule, with the chain rule on $\ln(\ln x)$ , inner function $\ln x$ : $\frac{y'}{y} = \ln(\ln x) + x \cdot \frac{1}{\ln x} \cdot \frac{1}{x} = \ln(\ln x) + \frac{1}{\ln x}$ . At $x = e$ : $\ln e = 1$ , so $\ln(\ln e) = \ln 1 = 0$ and $\frac{1}{\ln e} = 1$ ; also $y(e) = 1^e = 1$ . Hence $y'(e) = 1 \cdot (0 + 1) = 1$ . The two logarithms stacked are the trap: $\ln(\ln x)$ is not $(\ln x)^2$ and not $\ln x \cdot \ln x$ .

e)

No. The rule of thumb is about ONE power: here $x$ is the exponent of the factor $3^x$ and the base of the factor $x^3$ , and each factor has a rule of its own. Product rule: $y' = 3^x\ln 3 \cdot x^3 + 3^x \cdot 3x^2 = 3^x x^2(x\ln 3 + 3)$ , after factoring $3^x x^2$ . At $x = 1$ : $3(\ln 3 + 3) = 9 + 3\ln 3 \approx 12.2958$ . Logarithmic differentiation would also work, but it answers a question nobody asked: the decision is read on the SHAPE, variable in the base and the exponent of the same power, or not.

Exercise 7: The number e as a limit: a calculator table, then a proof

12 answers to check

Thomas 3.8 proves that $e = \lim_{x\to 0}(1 + x)^{1/x}$ , using only the derivative of $\ln$ at $1$ . The figure shows $y = (1 + x)^{1/x}$ : the function is not defined at $x = 0$ , and the hole is the limit.

A scientific calculator computes the values; it cannot compute the limit. A table suggests, a proof decides.

a) With a calculator, compute $(1 + x)^{1/x}$ to four decimals for $x = 0.1$ , $x = 0.01$ , $x = 0.001$ and $x = -0.01$ . What do the values suggest?
b) Write $\frac{d}{dx}\ln x\Big|_{x=1}$ as a limit, deduce $\lim_{h\to 0}\frac{\ln(1 + h)}{h}$ , and prove that $\lim_{x\to 0}(1 + x)^{1/x} = e$ .
c) Find $\lim_{x\to 0}\left(1 + \frac{x}{2}\right)^{3/x}$ and $\lim_{x\to 0}(1 - 3x)^{2/x}$ , each in the form $e^k$ .
d) Recognize each limit as a derivative at a point and evaluate it: $\lim_{h\to 0}\frac{5^h - 1}{h}$ and $\lim_{h\to 0}\frac{\log_3(9 + h) - 2}{h}$ .
e) Study $(1 + x)^{1/x^2}$ as $x \to 0^+$ and as $x \to 0^-$ . Does the limit as $x \to 0$ exist?

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Answers

a) $2.5937$ , $2.7048$ , $2.7169$ , $2.7320$ : the values approach $e \approx 2.7183$ , from below on the right and from above on the left
b) $\lim_{h\to 0}\frac{\ln(1 + h)}{h} = 1$ , so $\ln\left[(1 + x)^{1/x}\right] \to 1$ and $(1 + x)^{1/x} \to e$
c) $e^{3/2}$ and $e^{-6}$
d) $\ln 5 \approx 1.6094$ and $\frac{1}{9\ln 3} \approx 0.1011$
e) $+\infty$ from the right, $0$ from the left: the limit does not exist

a)

$1.1^{10} \approx 2.5937$ , $1.01^{100} \approx 2.7048$ , $1.001^{1000} \approx 2.7169$ and $0.99^{-100} \approx 2.7320$ , with the $y^x$ key. The values from the right increase toward $2.718\ldots$ , the value from the left is slightly above it: both sides seem to close in on $e \approx 2.7183$ . A table is evidence, not proof: it cannot rule out that the values turn away for $x = 10^{-9}$ , and a calculator pushed too far even returns $1$ , because it rounds $1 + 10^{-15}$ to $1$ . That is why b) is needed.

b)

By the definition of the derivative, $\frac{d}{dx}\ln x\Big|_{x=1} = \lim_{h\to 0}\frac{\ln(1 + h) - \ln 1}{h} = \lim_{h\to 0}\frac{\ln(1 + h)}{h}$ , and this derivative is known: $\frac{1}{1} = 1$ . So $\frac{\ln(1 + h)}{h} \to 1$ . The power law read from right to left turns this into the logarithm of our function: $\frac{1}{h}\ln(1 + h) = \ln\left[(1 + h)^{1/h}\right]$ . Hence $\ln\left[(1 + x)^{1/x}\right] \to 1$ as $x \to 0$ . Since $(1 + x)^{1/x} = e^{\ln[(1 + x)^{1/x}]}$ and the exponential is continuous, $(1 + x)^{1/x} \to e^1 = e$ . Stopping at $\ln[\ldots] \to 1$ and answering $1$ is the classic lost mark: that is the limit of the logarithm.

c)

$\ln\left[\left(1 + \frac{x}{2}\right)^{3/x}\right] = \frac{3}{x}\ln\left(1 + \frac{x}{2}\right) = \frac{3}{2} \cdot \frac{\ln(1 + u)}{u}$ with $u = \frac{x}{2} \to 0$ ; it tends to $\frac{3}{2}$ , so the limit is $e^{3/2}$ . The algebra is to multiply and divide by the right number: $\frac{3}{x} = \frac{3}{2} \cdot \frac{1}{u}$ . Similarly $\frac{2}{x}\ln(1 - 3x) = -6 \cdot \frac{\ln(1 + u)}{u}$ with $u = -3x \to 0$ : it tends to $-6$ , and the limit is $e^{-6}$ . The sign of $-3$ goes to the exponent; losing it gives $e^6$ , a huge number for a base below $1$ raised to large powers, which a calculator check at $x = 0.001$ exposes at once.

d)

Since $5^0 = 1$ , $\frac{5^h - 1}{h} = \frac{5^{0 + h} - 5^0}{h}$ is the difference quotient of $5^x$ at $0$ ; its limit is $\frac{d}{dx}5^x\Big|_{x=0} = 5^0\ln 5 = \ln 5 \approx 1.6094$ . Since $\log_3 9 = 2$ , $\frac{\log_3(9 + h) - 2}{h} = \frac{\log_3(9 + h) - \log_3 9}{h}$ is the difference quotient of $\log_3$ at $9$ , and its limit is $\frac{1}{9\ln 3} \approx 0.1011$ . The skill is to spot the three pieces: a function, a point, and the value at that point subtracted, here hidden as $1$ and as $2$ .

e)

$\ln\left[(1 + x)^{1/x^2}\right] = \frac{\ln(1 + x)}{x^2} = \frac{1}{x} \cdot \frac{\ln(1 + x)}{x}$ . The second factor tends to $1$ by b). As $x \to 0^+$ , $\frac{1}{x} \to +\infty$ , so the logarithm tends to $+\infty$ and the function to $+\infty$ . As $x \to 0^-$ , $\frac{1}{x} \to -\infty$ , the logarithm tends to $-\infty$ and the function to $0$ . The one-sided limits differ, so the limit does not exist. Calculator check: $1.01^{10000}$ is about $1.6 \times 10^{43}$ , $0.99^{10000}$ about $2 \times 10^{-44}$ . The form $1^\infty$ is the same as in b), and the outcome is completely different: the form decides nothing, the logarithm decides.

Exercise 8: Five statements to correct

10 answers to check

Each statement below was written on a MATH 203 assignment, and each is false. Say what is wrong, give the correct result, and check it on a value.

a) For $f(x) = 2x + \ln x$ , $(f^{-1})'(2) = \frac{1}{f'(2)} = \frac{2}{5}$ .
b) $\frac{d}{dx}3^x = x\,3^{x - 1}$ .
c) $\frac{d}{dx}\ln\left(e^{x^2}\right) = \frac{1}{e^{x^2}}$ .
d) $\ln(x - 1) - \ln(x + 1) = \frac{\ln(x - 1)}{\ln(x + 1)}$ , so its derivative follows from the quotient rule.
e) $\frac{d}{dx}\ln|x| = \frac{1}{|x|}$ .

Type your answers, the page tells you right or wrong 0/10

a)

The statement is

Correct

(f^{-1})'(2)

b)

The statement is

Correct derivative at

x = 1

c)

The statement is

Correct derivative at

x = 1

d)

The statement is

Correct derivative at

x = 3

e)

The statement is

Correct derivative at

x = -2

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Answers

a) False: $f(1) = 2$ , so $(f^{-1})'(2) = \frac{1}{f'(1)} = \frac{1}{3}$
b) False: $\frac{d}{dx}3^x = 3^x\ln 3$ , equal to $3\ln 3 \approx 3.2958$ at $x = 1$
c) False: $\ln(e^{x^2}) = x^2$ , derivative $2x$
d) False: the difference is $\ln\frac{x - 1}{x + 1}$ ; derivative $\frac{1}{x - 1} - \frac{1}{x + 1} = \frac{2}{x^2 - 1}$ , $\frac{1}{4}$ at $x = 3$
e) False: $\frac{d}{dx}\ln|x| = \frac{1}{x}$ , equal to $-\frac{1}{2}$ at $x = -2$

a)

FALSE. The theorem evaluates $f'$ at $a = f^{-1}(b)$ , not at $b$ . Here $f(1) = 2 + \ln 1 = 2$ , so $a = 1$ . $f'(x) = 2 + \frac{1}{x}$ , $f'(1) = 3$ , and $(f^{-1})'(2) = \frac{1}{3}$ . The number $\frac{1}{f'(2)} = \frac{1}{5/2} = \frac{2}{5}$ is the slope of $f^{-1}$ at $b = f(2) = 4 + \ln 2$ , another point. Correct: find $a$ first, write the pair $(a, b) = (1, 2)$ , then invert $f'(a)$ .

b)

FALSE. The power rule needs a constant exponent; in $3^x$ the exponent is the variable and the base is the constant. Rewrite $3^x = e^{x\ln 3}$ : its derivative is $e^{x\ln 3}\ln 3 = 3^x\ln 3$ . At $x = 1$ : $3\ln 3 \approx 3.2958$ , while the statement gives $1 \cdot 3^0 = 1$ . Another check: at $x = 0$ the statement gives slope $0$ , while $3^x$ crosses the $y$ axis rising with slope $\ln 3 > 0$ , as $2^x$ and $4^x$ do on the figure of exercise 4.

c)

FALSE twice. First, $\ln$ and $\exp$ undo each other: $\ln(e^{x^2}) = x^2$ , whose derivative is $2x$ . Second, even without simplifying, the chain rule gives $\frac{u'}{u}$ with $u = e^{x^2}$ , $u' = 2xe^{x^2}$ , so $\frac{2xe^{x^2}}{e^{x^2}} = 2x$ ; the statement forgot the derivative of the inside. At $x = 1$ : $2$ , not $\frac{1}{e}$ . Simplifying first leaves nothing to forget.

d)

FALSE. The law is $\ln a - \ln b = \ln\frac{a}{b}$ , the logarithm of a quotient, not a quotient of logarithms. And no rewriting is needed: differentiate the difference term by term, $\frac{1}{x - 1} - \frac{1}{x + 1} = \frac{(x + 1) - (x - 1)}{(x - 1)(x + 1)} = \frac{2}{x^2 - 1}$ , for $x > 1$ . At $x = 3$ : $\frac{2}{8} = \frac{1}{4}$ . Numerically, at $x = 3$ the difference is $\ln 2 - \ln 4 = -\ln 2 \approx -0.69$ while the quotient is $\frac{\ln 2}{\ln 4} = \frac{1}{2}$ : not even the same sign.

e)

FALSE. For $x < 0$ , $\ln|x| = \ln(-x)$ , and the chain rule with inner function $-x$ gives $\frac{-1}{-x} = \frac{1}{x}$ . So $\frac{d}{dx}\ln|x| = \frac{1}{x}$ for every $x \ne 0$ , negative on the left branch. At $x = -2$ : $-\frac{1}{2}$ , not $\frac{1}{2}$ . The graph agrees: as $x$ increases from $-5$ to $-1$ , $|x|$ decreases, so $\ln|x|$ decreases, and a decreasing function cannot have a positive slope there.

Exercise 9: Rates on logarithmic scales: decibels and pH

6 answers to check

The sound level of a sound of intensity $I$ , in W/m², is $\beta = 10\log_{10}\frac{I}{I_0}$ decibels, with $I_0 = 10^{-12}$ W/m². The pH of a solution is $\mathrm{pH} = -\log_{10}[\mathrm{H^+}]$ , the concentration in mol/L. Both are logarithms in base $10$ , so every rate on these scales carries the factor $\frac{1}{\ln 10}$ .

In open air, a loudspeaker produces at a distance of $r$ metres, $r \ge 1$ , the intensity $I(r) = \frac{10^{-2}}{r^2}$ W/m². The figure shows its sound level against the distance. Separately, $50$ mL of an acid solution with $[\mathrm{H^+}] = 0.01$ mol/L is diluted with $V$ mL of pure water; the number of moles of $\mathrm{H^+}$ does not change, so $[\mathrm{H^+}] = \frac{0.5}{50 + V}$ mol/L.

a) Using the laws of logarithms, show that $\beta(r) = 100 - 20\log_{10} r$ , and compute $\beta(10)$ .
b) Compute $\beta'(r)$ , then $\beta'(10)$ to four decimals, with its unit and the meaning of its sign.
c) At what distance does the sound level drop at a rate of exactly $1$ dB per metre? Give it to two decimals.
d) Show that $\mathrm{pH}(V) = \log_{10}(100 + 2V)$ , check $\mathrm{pH}(0) = 2$ , and compute the rate of change of the pH at $V = 0$ and at $V = 450$ mL. By what factor has the rate dropped?
e) For which volume of added water does the pH rise at the rate of $0.001$ pH unit per mL? Give it to two decimals.

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Answers

a) $\beta = 10\log_{10}\frac{10^{10}}{r^2} = 100 - 20\log_{10} r$ ; $\beta(10) = 80$ dB
b) $\beta'(r) = -\frac{20}{r\ln 10}$ ; $\beta'(10) = -\frac{2}{\ln 10} \approx -0.8686$ dB per metre: the level falls as the listener moves away
c) $r = \frac{20}{\ln 10} \approx 8.69$ m
d) $\mathrm{pH}'(V) = \frac{1}{(50 + V)\ln 10}$ ; $\approx 0.0087$ per mL at $V = 0$ , $\approx 0.00087$ at $V = 450$ : ten times smaller
e) $V = \frac{1000}{\ln 10} - 50 \approx 384.29$ mL

a)

$\frac{I}{I_0} = \frac{10^{-2}}{r^2 \cdot 10^{-12}} = \frac{10^{10}}{r^2}$ : the powers of ten combine first, $10^{-2}/10^{-12} = 10^{-2 + 12}$ , the exponent rule where copies most often go wrong. Then, for $r \ge 1$ , $\log_{10}\frac{10^{10}}{r^2} = \log_{10}10^{10} - \log_{10}r^2 = 10 - 2\log_{10}r$ , and $\beta(r) = 10(10 - 2\log_{10}r) = 100 - 20\log_{10}r$ . At $r = 10$ : $\beta = 100 - 20 = 80$ dB. Every tenfold increase of the distance costs $20$ dB, which is the shape of the curve on the figure: steep near the speaker, almost flat far away.

b)

$\log_{10}r = \frac{\ln r}{\ln 10}$ , so $\beta'(r) = -20 \cdot \frac{1}{r\ln 10} = -\frac{20}{r\ln 10}$ dB per metre. At $r = 10$ : $\beta'(10) = -\frac{2}{\ln 10} \approx -0.8686$ dB per metre. The sign says the level DECREASES as the distance increases, and the unit is dB per metre, the unit of $\beta$ over the unit of $r$ . Dropping the $\frac{1}{\ln 10}$ gives $-2$ dB per metre, more than twice the truth. Differentiating the original $10\log_{10}\frac{10^{-2}}{r^2 \cdot 10^{-12}}$ directly works too, through a quotient and a chain rule; the expansion of a) made it one line.

c)

The level drops at $1$ dB per metre when $\beta'(r) = -1$ : $\frac{20}{r\ln 10} = 1$ , so $r = \frac{20}{\ln 10} \approx 8.69$ m. Closer to the speaker the drop is faster, farther it is slower, since $|\beta'(r)|$ is inversely proportional to $r$ : at $5$ m it is $\frac{4}{\ln 10} \approx 1.74$ dB per metre, at $20$ m only $\frac{1}{\ln 10} \approx 0.43$ .

d)

$\mathrm{pH} = -\log_{10}\frac{0.5}{50 + V} = \log_{10}\frac{50 + V}{0.5} = \log_{10}(100 + 2V)$ , using $-\log_{10}\frac{a}{b} = \log_{10}\frac{b}{a}$ and $\frac{1}{0.5} = 2$ . At $V = 0$ : $\log_{10}100 = 2$ . Chain rule with inner function $100 + 2V$ : $\mathrm{pH}'(V) = \frac{2}{(100 + 2V)\ln 10} = \frac{1}{(50 + V)\ln 10}$ pH unit per mL. At $V = 0$ : $\frac{1}{50\ln 10} \approx 0.0087$ ; at $V = 450$ : $\frac{1}{500\ln 10} \approx 0.00087$ , where the pH is $\log_{10}1000 = 3$ . The rate has dropped by a factor of $10$ , the ratio of the total volumes $\frac{500}{50}$ : each new millilitre dilutes a larger volume and changes the concentration by a smaller fraction.

e)

$\frac{1}{(50 + V)\ln 10} = 0.001$ gives $50 + V = \frac{1000}{\ln 10} \approx 434.29$ , so $V \approx 384.29$ mL. Round only at the end: rounding $\ln 10$ to $2.3$ first gives $384.78$ , off by half a millilitre. At that moment the pH is $\log_{10}(100 + 768.59) \approx 2.94$ .

Exercise 10: A car, a line of credit, and the relative rate hidden in ln a

8 answers to check

A quantity $Q(t) = Q_0a^t$ changes at the rate $Q'(t) = Q_0a^t\ln a = (\ln a)\,Q(t)$ : its RELATIVE rate $\frac{Q'}{Q}$ is the constant $\ln a$ , not $a - 1$ . For $Q(t) = Q_0e^{rt}$ , continuous compounding, the relative rate is $r$ itself.

A car bought for $24\,000$ dollars loses $15\%$ of its value each year: $V(t) = 24\,000(0.85)^t$ dollars, $t$ in years. A line of credit of $6000$ dollars grows at $7.2\%$ compounded continuously: $B(t) = 6000e^{0.072t}$ dollars. Give rates in dollars per year, to the cent, and times to two decimals.

The figure below shows the value $V$ of the car.

a) Compute $V'(3)$ and the relative rate $\frac{V'}{V}$ to four decimals. Why is it not $-0.15$ ?
b) When is the car losing value at the rate of $1500$ dollars per year?
c) When is interest accruing on the line of credit at the rate of $600$ dollars per year?
d) Another lender quotes $7.2\%$ compounded ANNUALLY: $C(t) = 6000(1.072)^t$ . Give its relative rate to four decimals, and the annual rate, in percent to two decimals, that is equivalent to $7.2\%$ compounded continuously.
e) Find the doubling time of $B$ and of $C$ , and explain which number decides it.

Show the solution

Answers

a) $V'(3) = 24\,000(0.85)^3\ln 0.85 \approx -2395.37$ dollars per year; $\frac{V'}{V} = \ln 0.85 \approx -0.1625$ per year
b) $t = \frac{\ln\left(1500/(-24\,000\ln 0.85)\right)}{\ln 0.85} \approx 5.88$ years
c) $B = \frac{600}{0.072} = \frac{25\,000}{3}$ , so $t = \frac{\ln(25/18)}{0.072} \approx 4.56$ years
d) $\ln 1.072 \approx 0.0695$ per year; $e^{0.072} - 1 \approx 7.47\%$
e) $B$ : $\frac{\ln 2}{0.072} \approx 9.63$ years; $C$ : $\frac{\ln 2}{\ln 1.072} \approx 9.97$ years; the relative rate decides

a)

By the rule for $a^u$ : $V'(t) = 24\,000(0.85)^t\ln 0.85$ . At $t = 3$ : $(0.85)^3 = 0.614125$ , $V(3) = 14\,739$ dollars, and $V'(3) = 14\,739\ln 0.85 \approx -2395.37$ dollars per year. The sign says the car is LOSING value, since $\ln 0.85 < 0$ because $0.85 < 1$ . The relative rate is $\frac{V'}{V} = \ln 0.85 \approx -0.1625$ per year, at every $t$ . It is not $-0.15$ : the $15\%$ is the loss over one whole year, measured against the value at the start of the year, while $\ln 0.85$ is the instantaneous rate; since the value shrinks during the year, a larger instantaneous rate is needed to lose the same $15\%$ . The solution figure shows the tangent at $t = 3$ .

b)

$V'(t) = -1500$ : $24\,000(0.85)^t\ln 0.85 = -1500$ , so $(0.85)^t = \frac{1500}{-24\,000\ln 0.85} \approx 0.3846$ . Take $\ln$ of both sides and use the power law: $t\ln 0.85 = \ln 0.3846$ , so $t \approx \frac{-0.9556}{-0.1625} \approx 5.88$ years. Both logarithms are negative and their quotient is positive: a negative time here would mean a sign lost when the equation was divided by $\ln 0.85$ .

c)

$B'(t) = 0.072B(t)$ , so interest accrues at $600$ dollars per year when $B = \frac{600}{0.072} = \frac{25\,000}{3} \approx 8333.33$ dollars. Then $6000e^{0.072t} = \frac{25\,000}{3}$ gives $e^{0.072t} = \frac{25\,000}{18\,000} = \frac{25}{18}$ , and $t = \frac{\ln(25/18)}{0.072} \approx 4.56$ years. The rate in dollars per year grows with the balance; the relative rate stays $0.072$ .

d)

$C'(t) = 6000(1.072)^t\ln 1.072$ , so $\frac{C'}{C} = \ln 1.072 \approx 0.0695$ per year, a little LESS than $0.072$ : annual compounding at $7.2\%$ is slower than continuous compounding at $7.2\%$ . Conversely, in one year $B$ is multiplied by $e^{0.072} \approx 1.07466$ , so the equivalent annual rate is $e^{0.072} - 1 \approx 7.47\%$ . The quoted $7.2\%$ is the same number in both contracts; the relative rates, $0.072$ and $\ln 1.072$ , are not.

e)

$e^{0.072t} = 2$ gives $t = \frac{\ln 2}{0.072} \approx 9.63$ years; $(1.072)^t = 2$ gives $t = \frac{\ln 2}{\ln 1.072} \approx 9.97$ years. In both cases the doubling time is $\frac{\ln 2}{\text{relative rate}}$ , and that is the number that decides: $0.072$ for continuous compounding, $\ln 1.072 \approx 0.0695$ for annual compounding. A doubling time computed as $\frac{\ln 2}{0.072}$ for the annual contract confuses the quoted rate with the relative rate, the same confusion as $-0.15$ in a).

Corrected exercises: derivatives of inverse functions and logarithms (MATH 203)

Course recap

Part A: the basics (/50)

Exercise 1: The derivative of ln: the domain first, the inner function named

Exercise 2: The laws of logarithms BEFORE differentiating, and only the true ones

Exercise 3: The derivative of an inverse function at a point, without a formula for the inverse

Exercise 4: Logarithms in other bases: the constant ln a, even when the base moves

Exercise 5: Logarithmic differentiation of heavy products and quotients

Part B: problems and reasoning (/50)

Exercise 6: A variable in the base AND in the exponent of the same power

Exercise 7: The number e as a limit: a calculator table, then a proof

Exercise 8: Five statements to correct

Exercise 9: Rates on logarithmic scales: decibels and pH

Exercise 10: A car, a line of credit, and the relative rate hidden in ln a

See also

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