Exercise 1: Powers of a function: rewrite the exponent, then name u

The Chain Rule (Thomas 3.6): if $f(u)$ is differentiable at $u = g(x)$ and $g$ is differentiable at $x$ , then $(f \circ g)'(x) = f'(g(x))\, g'(x)$ . Thomas reads it as the Outside-Inside Rule: differentiate the outside function, evaluate it at the inside function left alone, then multiply by the derivative of the inside. Its most used case is the Power Chain Rule, $\frac{d}{dx}u^n = n u^{n-1}\frac{du}{dx}$ , valid for every real exponent $n$ .

In MATH 203 the marks on this exercise are not lost in the rule but in the algebra around it. Before differentiating, REWRITE: a root is a fractional exponent, a constant over a power is a negative exponent. Then compute the new exponent $n - 1$ with care ( $-3 - 1 = -4$ , $\frac{2}{3} - 1 = -\frac{1}{3}$ ). Begin every part with the sentence that names the inside function.

a) Differentiate $y = \frac{6}{(2x^2 + 1)^3}$ and give $y'(1)$ .
b) Differentiate $y = \sqrt[3]{(x^2 - 2x)^2}$ and give $y'(4)$ . Where is $y'$ undefined?
c) Differentiate $y = \left(\sqrt{x} - \frac{2}{x}\right)^3$ , rewriting the INSIDE as powers of $x$ first, and give $y'(4)$ .
d) Differentiate $y = \frac{4}{\sqrt{\sin x}}$ and give the exact value of $y'\left(\frac{\pi}{6}\right)$ .
e) A student writes $\frac{d}{dx}\frac{1}{(3x - 1)^2} = \frac{1}{2(3x - 1) \cdot 3}$ . Explain the error, give the correct derivative and its value at $x = 1$ .

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Answers

a) $y' = -\frac{72x}{(2x^2 + 1)^4}$ , $y'(1) = -\frac{8}{9}$
b) $y' = \frac{4(x - 1)}{3\sqrt[3]{x^2 - 2x}}$ , $y'(4) = 2$ ; undefined at $x = 0$ and $x = 2$
c) $y' = 3\left(\sqrt x - \frac{2}{x}\right)^2\left(\frac{1}{2\sqrt x} + \frac{2}{x^2}\right)$ , $y'(4) = \frac{81}{32}$
d) $y' = -\frac{2\cos x}{(\sin x)^{3/2}}$ , $y'\left(\frac{\pi}{6}\right) = -2\sqrt 6 \approx -4.8990$
e) $\frac{d}{dx}(3x - 1)^{-2} = -\frac{6}{(3x - 1)^3}$ , which is $-\frac{3}{4}$ at $x = 1$

a) Rewrite first: $y = 6(2x^2 + 1)^{-3}$ , the $6$ staying upstairs. Let $u = 2x^2 + 1$ , so $y = 6u^{-3}$ , $\frac{dy}{du} = -18u^{-4}$ , and $u' = 4x$ . By the Power Chain Rule, $y' = -18(2x^2 + 1)^{-4} \cdot 4x = -\frac{72x}{(2x^2 + 1)^4}$ . The exponent is $-3 - 1 = -4$ : moving TOWARD zero, to $-2$ , is the classic slip with a negative exponent. At $x = 1$ : $u = 3$ , $3^4 = 81$ , so $y'(1) = -\frac{72}{81} = -\frac{8}{9}$ . The quotient rule gives the same result, but with a chain rule hidden in the derivative of the denominator and a fraction to simplify at the end.

b) The index of the root goes in the denominator of the exponent: $\sqrt[3]{(x^2 - 2x)^2} = (x^2 - 2x)^{2/3}$ . Let $u = x^2 - 2x$ , $u' = 2x - 2$ . Then $y' = \frac{2}{3}u^{-1/3} u' = \frac{2(2x - 2)}{3(x^2 - 2x)^{1/3}} = \frac{4(x - 1)}{3\sqrt[3]{x^2 - 2x}}$ , since $\frac{2}{3} - 1 = -\frac{1}{3}$ and a negative exponent sends the cube root to the denominator. At $x = 4$ : $u = 8$ , $\sqrt[3]{8} = 2$ , so $y'(4) = \frac{4 \cdot 3}{3 \cdot 2} = 2$ . The function is defined everywhere, but $y'$ is undefined where $u = 0$ , at $x = 0$ and $x = 2$ : the rewriting $u^{-1/3}$ is what shows it.

c) Rewrite the inside: $u = x^{1/2} - 2x^{-1}$ , so $u' = \frac{1}{2}x^{-1/2} + 2x^{-2} = \frac{1}{2\sqrt x} + \frac{2}{x^2}$ . The minus sign of $-2x^{-1}$ meets the exponent $-1$ and becomes a PLUS: that sign is where the marks go. The outside function is the cube: $y' = 3u^2 u' = 3\left(\sqrt x - \frac{2}{x}\right)^2\left(\frac{1}{2\sqrt x} + \frac{2}{x^2}\right)$ . At $x = 4$ : $u = 2 - \frac{1}{2} = \frac{3}{2}$ and $u' = \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$ , so $y'(4) = 3 \cdot \frac{9}{4} \cdot \frac{3}{8} = \frac{81}{32}$ . Expanding the cube first would produce four terms with half-integer exponents: four chances to lose a sign for no gain.

d) Rewrite: $y = 4(\sin x)^{-1/2}$ . The outside function is $4u^{-1/2}$ , of derivative $-2u^{-3/2}$ (since $-\frac{1}{2} - 1 = -\frac{3}{2}$ ); the inside is $u = \sin x$ , with $u' = \cos x$ . So $y' = -2(\sin x)^{-3/2}\cos x = -\frac{2\cos x}{(\sin x)^{3/2}}$ . At $\frac{\pi}{6}$ : $\sin\frac{\pi}{6} = \frac{1}{2}$ and $\left(\frac{1}{2}\right)^{-3/2} = 2^{3/2} = 2\sqrt 2$ , a negative exponent flipping the fraction; $\cos\frac{\pi}{6} = \frac{\sqrt 3}{2}$ . So $y'\left(\frac{\pi}{6}\right) = -2 \cdot 2\sqrt 2 \cdot \frac{\sqrt 3}{2} = -2\sqrt 6 \approx -4.8990$ . The calculator confirms the decimal; the exact value is the expected answer.

e) The student took the derivative of $\frac{1}{v}$ to be $\frac{1}{v'}$ , which is no rule at all, and then differentiated the square as if it were the function. Rewrite: $y = (3x - 1)^{-2}$ , with $u = 3x - 1$ and $u' = 3$ . Then $y' = -2(3x - 1)^{-3} \cdot 3 = -\frac{6}{(3x - 1)^3}$ . At $x = 1$ : $-\frac{6}{8} = -\frac{3}{4}$ , while the student's expression gives $\frac{1}{12}$ , a POSITIVE slope. A sign check exposes it: for $x > \frac{1}{3}$ the denominator $(3x - 1)^2$ grows, so $y$ decreases and its derivative must be negative. A constant over a power is a negative power; rewriting it removes the temptation.

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Exercise 2: Three and four layers: one factor per layer

When a function is built in several layers, the chain rule is applied once per layer, and the derivative is a PRODUCT with one factor per layer. Thomas calls this repeated use of the Chain Rule: $\frac{d}{dx}f(g(h(x))) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$ .

The safe method: list the layers from the outside in BEFORE writing anything, then write one factor per layer, each outside derivative keeping everything inside it untouched, then count the factors against the layers. A calculator helps with the final decimal only if it is in RADIAN mode.

a) Differentiate $y = \sqrt{1 + \sqrt x}$ and give $y'(9)$ .
b) Differentiate $y = \frac{1}{\sqrt{1 + \cos 2x}}$ and give the exact value of $y'\left(\frac{\pi}{3}\right)$ .
c) Differentiate $y = \cos\left(\sqrt{x^2 + 1}\right)$ and give $y'(\sqrt 3)$ to four decimals.
d) Let $h(t) = e^{\sin^2(\pi t)}$ . List its four layers, find $h'(t)$ and the exact value of $h'\left(\frac{1}{4}\right)$ .
e) A student writes $\frac{d}{dx}(3 + \sin 3x)^{-2} = -2(3 + \sin 3x)^{-3}\cos 3x$ . Count the layers, correct the answer and give its value at $x = 0$ .

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Answers

a) $y' = \frac{1}{4\sqrt x\,\sqrt{1 + \sqrt x}}$ , $y'(9) = \frac{1}{24}$
b) $y' = \frac{\sin 2x}{(1 + \cos 2x)^{3/2}}$ , $y'\left(\frac{\pi}{3}\right) = \sqrt 6$
c) $y' = -\frac{x\sin\sqrt{x^2 + 1}}{\sqrt{x^2 + 1}}$ , $y'(\sqrt 3) = -\frac{\sqrt 3}{2}\sin 2 \approx -0.7875$
d) Layers $e^{(\cdot)}$ , square, sine, $\pi t$ ; $h'(t) = 2\pi\sin(\pi t)\cos(\pi t)\, e^{\sin^2(\pi t)}$ , $h'\left(\frac{1}{4}\right) = \pi\sqrt e$
e) Three layers; $y' = -\frac{6\cos 3x}{(3 + \sin 3x)^3}$ , which is $-\frac{2}{9}$ at $x = 0$

a) Layers: the outer square root, then $1 + w$ , then $w = \sqrt x$ . Factors: $\frac{1}{2\sqrt{1 + \sqrt x}}$ for the outer root, $1$ for $1 + w$ , and $\frac{1}{2\sqrt x}$ for the inner root. So $y' = \frac{1}{2\sqrt{1 + \sqrt x}} \cdot \frac{1}{2\sqrt x} = \frac{1}{4\sqrt x\,\sqrt{1 + \sqrt x}}$ . At $x = 9$ : $\sqrt 9 = 3$ and $\sqrt 4 = 2$ , so $y'(9) = \frac{1}{4 \cdot 3 \cdot 2} = \frac{1}{24}$ . The two roots are two different layers, each with its own factor: the frequent error writes $\frac{1}{2\sqrt{1 + x}}$ , merging them.

b) Rewrite: $y = (1 + \cos 2x)^{-1/2}$ . Layers: the power $-\frac{1}{2}$ , then $1 + \cos w$ , then $w = 2x$ . Factors: $-\frac{1}{2}(1 + \cos 2x)^{-3/2}$ , then $-\sin 2x$ , then $2$ . The product: $y' = -\frac{1}{2} \cdot (-\sin 2x) \cdot 2 \cdot (1 + \cos 2x)^{-3/2} = \frac{\sin 2x}{(1 + \cos 2x)^{3/2}}$ . Two minus signs meet and cancel, one from the exponent and one from the cosine: losing either one flips every slope. At $\frac{\pi}{3}$ : $\sin\frac{2\pi}{3} = \frac{\sqrt 3}{2}$ and $1 + \cos\frac{2\pi}{3} = \frac{1}{2}$ , with $\left(\frac{1}{2}\right)^{3/2} = \frac{1}{2\sqrt 2}$ , so $y' = \frac{\sqrt 3}{2} \cdot 2\sqrt 2 = \sqrt 6$ . Check by an identity: $1 + \cos 2x = 2\cos^2 x$ , so on $\left(0, \frac{\pi}{2}\right)$ , $y = \frac{1}{\sqrt 2\cos x}$ and $y' = \frac{\sin x}{\sqrt 2\cos^2 x}$ , which is $\frac{\sqrt 3/2}{\sqrt 2/4} = \sqrt 6$ at $\frac{\pi}{3}$ as well.

c) Layers: $\cos$ , then the square root, then $x^2 + 1$ . Factors: $-\sin\sqrt{x^2 + 1}$ , then $\frac{1}{2\sqrt{x^2 + 1}}$ , then $2x$ . So $y' = -\frac{x\sin\sqrt{x^2 + 1}}{\sqrt{x^2 + 1}}$ , the $2$ cancelling. At $x = \sqrt 3$ : $\sqrt{x^2 + 1} = 2$ , so $y'(\sqrt 3) = -\frac{\sqrt 3}{2}\sin 2 \approx -0.7875$ , with $\sin 2$ meaning the sine of $2$ RADIANS. A calculator left in degree mode gives $\sin 2^\circ$ and the answer $-0.0302$ : every formula of the course assumes radians.

d) Layers, from the outside in: $e^{(\cdot)}$ , the square, the sine, and $\pi t$ . Four layers, four factors: $h'(t) = e^{\sin^2(\pi t)} \cdot 2\sin(\pi t) \cdot \cos(\pi t) \cdot \pi$ . The exponential comes back unchanged, with its whole exponent. At $t = \frac{1}{4}$ : $\sin\frac{\pi}{4} = \cos\frac{\pi}{4} = \frac{\sqrt 2}{2}$ , so $\sin^2 = \frac{1}{2}$ and $2\sin\cos = 1$ ; hence $h'\left(\frac{1}{4}\right) = e^{1/2} \cdot 1 \cdot \pi = \pi\sqrt e \approx 5.1797$ . The double-angle identity of Thomas 1.3 gives a shorter form, $h'(t) = \pi\sin(2\pi t)\, e^{\sin^2(\pi t)}$ .

e) Layers: the power $-2$ , then $3 + \sin w$ , then $w = 3x$ . Three layers, and the student wrote two factors: the inner $3$ is missing. Correct: $y' = -2(3 + \sin 3x)^{-3} \cdot \cos 3x \cdot 3 = -\frac{6\cos 3x}{(3 + \sin 3x)^3}$ . At $x = 0$ : $-\frac{6}{27} = -\frac{2}{9}$ , three times the student's $-\frac{2}{27}$ . The innermost factor is forgotten precisely because it is only a number; counting layers against factors catches it every time.

Exercise 3: The chain rule from a table of values

On a MATH 203 exam and in WeBWorK, the chain rule is often tested on functions known only through a table. The rule is then pure bookkeeping: $(f \circ g)'(a) = f'(g(a))\, g'(a)$ . Compute the inside value $g(a)$ FIRST, write it down as a number, and only then go to the row of that number to read $f'$ .

The table gives values of two differentiable functions $f$ and $g$ and of their derivatives. Some of the questions below cannot be answered from it: saying so, with the reason, is the answer.

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
$0$	$2$	$-1$	$1$	$3$
$1$	$4$	$3$	$4$	$-2$
$2$	$1$	$5$	$0$	$\frac{1}{2}$
$4$	$3$	$-2$	$2$	$6$

a) Let $h(x) = f(g(x))$ . Find $h'(0)$ and $h'(2)$ . A student answers $h'(0) = -3$ : what did the student compute?
b) Let $k(x) = g(f(x))$ . Find $k'(2)$ .
c) Let $P(x) = \sqrt{f(x)}$ and $Q(x) = g(\sqrt x)$ . Find $P'(1)$ and $Q'(4)$ .
d) Let $R(x) = f(g(f(x)))$ and $S(x) = 2^{g(x)}$ . Find $R'(0)$ , then the exact value of $S'(1)$ .
e) Among $(f \circ g)'(4)$ , $(g \circ f)'(4)$ and $(f \circ f)'(0)$ , compute those the table allows, and explain why one of them cannot be found.

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Answers

a) $h'(0) = f'(1)\,g'(0) = 9$ , $h'(2) = f'(0)\,g'(2) = -\frac{1}{2}$ ; $-3 = f'(0)\,g'(0)$ is the wrong row.
b) $k'(2) = g'(f(2))\,f'(2) = g'(1) \cdot 5 = -10$
c) $P'(1) = \frac{f'(1)}{2\sqrt{f(1)}} = \frac{3}{4}$ ; $Q'(4) = g'(2) \cdot \frac{1}{4} = \frac{1}{8}$
d) $R'(0) = f'(0)\,g'(2)\,f'(0) = \frac{1}{2}$ ; $S'(1) = 2^{4}\ln 2 \cdot (-2) = -32\ln 2$
e) $(f \circ g)'(4) = 30$ , $(f \circ f)'(0) = -5$ ; $(g \circ f)'(4)$ needs $g'(3)$ , not in the table.

a) $h'(0) = f'(g(0)) \cdot g'(0)$ . Read $g(0) = 1$ first, then $f'$ in the row $x = 1$ : $f'(1) = 3$ . With $g'(0) = 3$ , $h'(0) = 3 \cdot 3 = 9$ . The student's $-3$ is $f'(0) \cdot g'(0) = (-1) \cdot 3$ : the outside derivative was read at $x = 0$ instead of at $g(0) = 1$ . The outside function never sees $x$ , only the value that $g$ hands it. Similarly $h'(2) = f'(g(2)) \cdot g'(2) = f'(0) \cdot \frac{1}{2} = -\frac{1}{2}$ , since $g(2) = 0$ .

b) $k'(2) = g'(f(2)) \cdot f'(2)$ . First $f(2) = 1$ , then $g'(1) = -2$ ; with $f'(2) = 5$ , $k'(2) = -2 \cdot 5 = -10$ . The order of composition decides which function is read where: in $g \circ f$ , the table of $g'$ is read at a value of $f$ .

c) $P = \sqrt u$ with $u = f(x)$ , so $P'(x) = \frac{f'(x)}{2\sqrt{f(x)}}$ and $P'(1) = \frac{3}{2\sqrt 4} = \frac{3}{4}$ . For $Q(x) = g(\sqrt x)$ , the inside is $\sqrt x$ , of derivative $\frac{1}{2\sqrt x}$ : $Q'(x) = g'(\sqrt x) \cdot \frac{1}{2\sqrt x}$ . At $x = 4$ : $\sqrt 4 = 2$ , $g'(2) = \frac{1}{2}$ and $\frac{1}{2\sqrt 4} = \frac{1}{4}$ , so $Q'(4) = \frac{1}{8}$ . Reading $g'(4) = 6$ because the question says $x = 4$ gives $\frac{3}{2}$ , twelve times too large.

d) Three layers, three factors, each read at its own point. Follow the values from the inside: $f(0) = 2$ , then $g(2) = 0$ . So $R'(0) = f'(g(f(0))) \cdot g'(f(0)) \cdot f'(0) = f'(0) \cdot g'(2) \cdot f'(0) = (-1) \cdot \frac{1}{2} \cdot (-1) = \frac{1}{2}$ . The two factors $f'(0)$ are read at different stages that happen to share the value $0$ . For $S(x) = 2^{g(x)} = e^{g(x)\ln 2}$ , with $\ln 2$ a constant: $S'(x) = 2^{g(x)}\ln 2 \cdot g'(x)$ . At $x = 1$ : $g(1) = 4$ , so $S'(1) = 2^4 \ln 2 \cdot (-2) = -32\ln 2 \approx -22.1807$ .

e) $(f \circ g)'(4) = f'(g(4)) \cdot g'(4) = f'(2) \cdot 6 = 5 \cdot 6 = 30$ . $(f \circ f)'(0) = f'(f(0)) \cdot f'(0) = f'(2) \cdot (-1) = -5$ . But $(g \circ f)'(4) = g'(f(4)) \cdot f'(4) = g'(3) \cdot (-2)$ , and the table has no row $x = 3$ : the value cannot be found. Taking $g'(4) = 6$ instead would be the wrong-row error again, disguised as an answer.

Exercise 4: Exponentials in base a: the laws of exponents first

For $a > 0$ , $a^x = e^{x\ln a}$ , where $\ln a$ is a CONSTANT. With the inside function $u = x\ln a$ , the chain rule gives $\frac{d}{dx}a^x = a^x\ln a$ , and more generally $\frac{d}{dx}a^u = a^u \ln a \cdot u'$ . Nothing about logarithms is differentiated here.

Most of the work, and most of the lost marks, come BEFORE the derivative: rewrite with the laws of exponents so that a single exponential remains. $\frac{1}{a^x} = a^{-x}$ , $\sqrt{a^x} = a^{x/2}$ , $(a^x)^k = a^{kx}$ . The figure shows the curve of part d).

a) Differentiate $y = \frac{5}{2^x}$ and give $y'(1)$ .
b) Differentiate $y = \sqrt{3^x}$ and give $y'(2)$ .
c) Differentiate $y = 2^{x^2 - 2x}$ . Find the point where its tangent is horizontal, and give $y'(2)$ .
d) Differentiate $y = x^2\,3^{-x}$ , write $y'$ as a product of factors, and find the $x$ -values where the tangent is horizontal.
e) Find the base $a > 0$ for which $y = a^x$ satisfies $y' = 3y$ for all $x$ . Then find $a$ so that $y = a^{-2x}$ satisfies $y' = -y$ .

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Answers

a) $y' = -5\ln 2 \cdot 2^{-x}$ , $y'(1) = -\frac{5}{2}\ln 2 \approx -1.7329$
b) $y' = \frac{\ln 3}{2}\,3^{x/2}$ , $y'(2) = \frac{3}{2}\ln 3 \approx 1.6479$
c) $y' = (2x - 2)\ln 2 \cdot 2^{x^2 - 2x}$ ; horizontal at $\left(1, \frac{1}{2}\right)$ ; $y'(2) = 2\ln 2 \approx 1.3863$
d) $y' = x\,3^{-x}(2 - x\ln 3)$ ; horizontal at $x = 0$ and $x = \frac{2}{\ln 3} \approx 1.8205$
e) $a = e^3 \approx 20.0855$ ; $a = \sqrt e \approx 1.6487$

a) Rewrite: $y = 5 \cdot 2^{-x} = 5e^{-x\ln 2}$ . The inside function is $u = -x\ln 2$ , with $u' = -\ln 2$ . So $y' = 5 \cdot 2^{-x} \cdot (-\ln 2) = -5\ln 2 \cdot 2^{-x}$ . At $x = 1$ : $y'(1) = -\frac{5\ln 2}{2} \approx -1.7329$ . A student who applies the power rule to $2^{-x}$ and writes $-x\,2^{-x-1}$ confuses a variable exponent with a variable base: the power rule needs a constant exponent.

b) A power of a power multiplies the exponents: $\sqrt{3^x} = (3^x)^{1/2} = 3^{x/2}$ . With $u = \frac{x}{2}$ : $y' = 3^{x/2}\ln 3 \cdot \frac{1}{2}$ . At $x = 2$ : $y'(2) = \frac{3\ln 3}{2} \approx 1.6479$ . The Power Chain Rule on the square root reaches the same place with more writing: $\frac{1}{2}(3^x)^{-1/2} \cdot 3^x\ln 3 = \frac{\ln 3}{2}\,3^{x/2}$ , after using $3^x \cdot 3^{-x/2} = 3^{x/2}$ , another law of exponents.

c) The outside function is $2^u$ , the inside $u = x^2 - 2x$ : $y' = 2^{x^2 - 2x}\ln 2 \cdot (2x - 2)$ . The factor $2^{x^2 - 2x}$ is never zero and $\ln 2 \ne 0$ , so $y' = 0$ only when $2x - 2 = 0$ : $x = 1$ , where $y = 2^{-1} = \frac{1}{2}$ . The horizontal tangent is at $\left(1, \frac{1}{2}\right)$ . At $x = 2$ : $u = 0$ , so $y'(2) = 1 \cdot \ln 2 \cdot 2 = 2\ln 2 \approx 1.3863$ . Saying WHY the exponential factor cannot vanish is part of the answer.

d) The last operation is a product: product rule, with the chain rule on $3^{-x}$ , whose derivative is $3^{-x}\ln 3 \cdot (-1)$ . So $y' = 2x\,3^{-x} - x^2\ln 3 \cdot 3^{-x}$ . Factor the common factors, $x$ (its lowest power) and $3^{-x}$ : $y' = x\,3^{-x}(2 - x\ln 3)$ . Since $3^{-x} > 0$ , $y' = 0$ exactly when $x = 0$ or $x = \frac{2}{\ln 3} \approx 1.8205$ : two horizontal tangents, as the figure shows. At the second one, $3^{-2/\ln 3} = e^{-2}$ (because $3^{-x} = e^{-x\ln 3}$ ), so $y = \frac{4}{(\ln 3)^2}e^{-2} \approx 0.4485$ . The factored form makes the equation solvable in one line; the expanded form hides it.

e) By the chain rule, $(a^x)' = a^x\ln a = (\ln a)\, y$ . So $y' = 3y$ for all $x$ exactly when $\ln a = 3$ , that is $a = e^3 \approx 20.0855$ . For $y = a^{-2x}$ : $y' = a^{-2x}\ln a \cdot (-2) = -2\ln a \cdot y$ , and $-2\ln a = -1$ gives $\ln a = \frac{1}{2}$ , $a = e^{1/2} = \sqrt e \approx 1.6487$ . Every exponential $a^x$ is $e^{kx}$ with $k = \ln a$ , and the inside derivative $k$ is the factor that the derivative carries.

Exercise 5: Factor the lowest power: the algebra after the chain rule

When the product or quotient rule meets the chain rule, the derivative comes out as a SUM of terms containing the same brackets at different powers. It is not finished: a sum cannot be set equal to zero or signed. Factor each bracket at its LOWEST power, the most negative exponent ( $-\frac{1}{2}$ is lower than $\frac{1}{2}$ , and $-3$ is lower than $-2$ ). What is left inside the square bracket then has only non-negative powers and simplifies to a polynomial.

The rule for the leftover exponents: after taking out $u^{m}$ , a term containing $u^{n}$ keeps $u^{n - m}$ . Rewriting a quotient as a product with a negative exponent, before differentiating, usually saves the quotient rule altogether. The figure shows the curve of parts d) and e).

a) Differentiate $y = x\sqrt{1 - 2x}$ , write $y'$ as a single fraction, give $y'(-4)$ and the $x$ -value of the horizontal tangent.
b) Differentiate $y = \frac{(x + 2)^3}{(2x - 1)^2}$ by rewriting it as a product. Give $y'$ factored, $y'(1)$ , and the $x$ -values of the horizontal tangents.
c) Differentiate $y = \frac{x}{(x^2 + 4)^{3/2}}$ and show that $y' = \frac{4 - 2x^2}{(x^2 + 4)^{5/2}}$ . Give $y'(0)$ and the positive zero of $y'$ .
d) Differentiate $y = (x + 2)\sqrt[3]{x^2 - 1}$ , factor, and give $y'(3)$ .
e) Using d), find the $x$ -values where the curve has a horizontal tangent, and those where $y'$ does not exist. Compare with the figure.

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Answers

a) $y' = \frac{1 - 3x}{\sqrt{1 - 2x}}$ , $y'(-4) = \frac{13}{3}$ , horizontal at $x = \frac{1}{3}$
b) $y' = \frac{(x + 2)^2(2x - 11)}{(2x - 1)^3}$ , $y'(1) = -81$ , horizontal at $x = -2$ and $x = \frac{11}{2}$
c) $y' = \frac{4 - 2x^2}{(x^2 + 4)^{5/2}}$ , $y'(0) = \frac{1}{8}$ , positive zero $x = \sqrt 2$
d) $y' = \frac{5x^2 + 4x - 3}{3(x^2 - 1)^{2/3}}$ , $y'(3) = \frac{9}{2}$
e) Horizontal at $x = \frac{-2 \pm \sqrt{19}}{5}$ ( $\approx -1.2718$ and $0.4718$ ); $y'$ undefined at $x = \pm 1$ (vertical tangents)

a) $y = x(1 - 2x)^{1/2}$ , a product. With $u = 1 - 2x$ , $u' = -2$ : $y' = (1 - 2x)^{1/2} + x \cdot \frac{1}{2}(1 - 2x)^{-1/2}(-2) = (1 - 2x)^{1/2} - x(1 - 2x)^{-1/2}$ . The lowest power of the bracket is $-\frac{1}{2}$ ; taking it out of $(1 - 2x)^{1/2}$ leaves $(1 - 2x)^{1/2 - (-1/2)} = (1 - 2x)^1$ . So $y' = (1 - 2x)^{-1/2}\left[(1 - 2x) - x\right] = \frac{1 - 3x}{\sqrt{1 - 2x}}$ , for $x < \frac{1}{2}$ . At $x = -4$ : $\frac{13}{\sqrt 9} = \frac{13}{3}$ . Horizontal tangent where the numerator vanishes: $x = \frac{1}{3}$ .

b) Rewrite: $y = (x + 2)^3(2x - 1)^{-2}$ . Product rule, chain rule on each factor: $y' = 3(x + 2)^2(2x - 1)^{-2} + (x + 2)^3 \cdot (-2)(2x - 1)^{-3} \cdot 2 = 3(x + 2)^2(2x - 1)^{-2} - 4(x + 2)^3(2x - 1)^{-3}$ . Lowest powers: $(x + 2)^2$ and $(2x - 1)^{-3}$ , NOT $(2x - 1)^{-2}$ . The leftover: $y' = (x + 2)^2(2x - 1)^{-3}\left[3(2x - 1) - 4(x + 2)\right] = \frac{(x + 2)^2(2x - 11)}{(2x - 1)^3}$ . At $x = 1$ : $\frac{9 \cdot (-9)}{1} = -81$ . Horizontal tangents at $x = -2$ and $x = \frac{11}{2}$ . Factoring $(2x - 1)^{-2}$ instead would leave $(2x - 1)^{-1}$ inside the bracket, a fraction inside the factorisation: the sign that the wrong power was taken out.

c) Rewrite: $y = x(x^2 + 4)^{-3/2}$ . Then $y' = (x^2 + 4)^{-3/2} + x \cdot \left(-\frac{3}{2}\right)(x^2 + 4)^{-5/2} \cdot 2x = (x^2 + 4)^{-3/2} - 3x^2(x^2 + 4)^{-5/2}$ . The lowest power is $-\frac{5}{2}$ , and $-\frac{3}{2} - \left(-\frac{5}{2}\right) = 1$ : $y' = (x^2 + 4)^{-5/2}\left[(x^2 + 4) - 3x^2\right] = \frac{4 - 2x^2}{(x^2 + 4)^{5/2}}$ . At $x = 0$ : $\frac{4}{4^{5/2}} = \frac{4}{32} = \frac{1}{8}$ . Zeros: $x^2 = 2$ , the positive one is $\sqrt 2 \approx 1.4142$ . The quotient rule gives $\frac{(x^2 + 4)^{3/2} - 3x^2(x^2 + 4)^{1/2}}{(x^2 + 4)^3}$ , a complex fraction that needs this very factorisation to be finished.

d) $y = (x + 2)(x^2 - 1)^{1/3}$ . With $u = x^2 - 1$ : $y' = (x^2 - 1)^{1/3} + (x + 2) \cdot \frac{1}{3}(x^2 - 1)^{-2/3} \cdot 2x$ . Lowest power $-\frac{2}{3}$ , and $\frac{1}{3} - \left(-\frac{2}{3}\right) = 1$ : $y' = (x^2 - 1)^{-2/3}\left[(x^2 - 1) + \frac{2x(x + 2)}{3}\right] = \frac{3x^2 - 3 + 2x^2 + 4x}{3(x^2 - 1)^{2/3}} = \frac{5x^2 + 4x - 3}{3(x^2 - 1)^{2/3}}$ . At $x = 3$ : $x^2 - 1 = 8$ , $8^{2/3} = 4$ , so $y'(3) = \frac{45 + 12 - 3}{12} = \frac{54}{12} = \frac{9}{2}$ .

e) $y' = 0$ when $5x^2 + 4x - 3 = 0$ : discriminant $16 + 60 = 76$ , so $x = \frac{-4 \pm \sqrt{76}}{10} = \frac{-2 \pm \sqrt{19}}{5}$ , about $-1.2718$ and $0.4718$ . The denominator vanishes at $x = \pm 1$ , where $y'$ does not exist: the numerator is $6$ at $x = 1$ and $-2$ at $x = -1$ , not zero, so $|y'|$ grows without bound and the curve has a VERTICAL tangent there, as the figure shows at the two marked points. The factored form answers both questions at once; the unfactored sum answers neither.

Exercise 6: Second and higher derivatives: the chain inside the product

The first derivative of a composite is a PRODUCT, $f'(g(x)) \cdot g'(x)$ , so the second derivative needs the product rule, and the chain rule again inside the first factor. Nothing new is required, only the discipline of seeing the product that the first derivative created.

Write $y'$ with negative exponents rather than as a quotient before differentiating again, and finish $y''$ by factoring the lowest power: that is where the marks of this exercise are decided.

a) Let $y = \frac{1}{1 + x^2}$ . Show that $y'' = \frac{6x^2 - 2}{(1 + x^2)^3}$ , give $y''(1)$ and the positive zero of $y''$ .
b) Let $y = \tan(2x)$ . Find $y''$ and its value at $x = \frac{\pi}{8}$ .
c) Let $y = (4 - x^2)^{3/2}$ . Find $y''$ as a single fraction and give $y''(1)$ exactly.
d) Let $g(x) = f(x^2)$ , where $f$ is twice differentiable with $f'(4) = -1$ and $f''(4) = 3$ . Express $g''(x)$ in terms of $f'$ and $f''$ and find $g''(2)$ . A student answers $48$ : what was forgotten?
e) Let $y = (1 - 2x)^{-1}$ . Find $y'$ , $y''$ , $y'''$ , guess the formula for the $n$ -th derivative, and give $y'''(0)$ and $y^{(5)}(0)$ .

Show the solution

Answers

a) $y'' = \frac{6x^2 - 2}{(1 + x^2)^3}$ , $y''(1) = \frac{1}{2}$ , positive zero $x = \frac{1}{\sqrt 3}$
b) $y'' = 8\sec^2(2x)\tan(2x)$ , $y''\left(\frac{\pi}{8}\right) = 16$
c) $y'' = \frac{6x^2 - 12}{\sqrt{4 - x^2}}$ , $y''(1) = -2\sqrt 3$
d) $g''(x) = 2f'(x^2) + 4x^2 f''(x^2)$ , $g''(2) = 46$ ; $48$ forgets the term $2f'(x^2)$ .
e) $y^{(n)} = 2^n\, n!\,(1 - 2x)^{-(n+1)}$ ; $y'''(0) = 48$ , $y^{(5)}(0) = 3840$

a) $y = (1 + x^2)^{-1}$ , so $y' = -(1 + x^2)^{-2} \cdot 2x = -2x(1 + x^2)^{-2}$ , a product. Product rule, chain rule on the second factor: $y'' = -2(1 + x^2)^{-2} + (-2x) \cdot (-2)(1 + x^2)^{-3} \cdot 2x = -2(1 + x^2)^{-2} + 8x^2(1 + x^2)^{-3}$ . Lowest power $-3$ : $y'' = (1 + x^2)^{-3}\left[-2(1 + x^2) + 8x^2\right] = \frac{6x^2 - 2}{(1 + x^2)^3}$ . At $x = 1$ : $\frac{4}{8} = \frac{1}{2}$ . Zeros: $x^2 = \frac{1}{3}$ , the positive one $\frac{1}{\sqrt 3} \approx 0.5774$ . Differentiating $\frac{-2x}{(1 + x^2)^2}$ by the quotient rule works too, but produces a factor $(1 + x^2)$ to cancel against $(1 + x^2)^4$ : one more place to slip.

b) $y' = \sec^2(2x) \cdot 2 = 2(\sec 2x)^2$ . For the second derivative, $\sec^2(2x)$ is itself three layers: the square, the secant, $2x$ . So $y'' = 2 \cdot 2\sec(2x) \cdot \sec(2x)\tan(2x) \cdot 2 = 8\sec^2(2x)\tan(2x)$ . At $\frac{\pi}{8}$ : $2x = \frac{\pi}{4}$ , $\sec^2\frac{\pi}{4} = 2$ and $\tan\frac{\pi}{4} = 1$ , so $y''\left(\frac{\pi}{8}\right) = 16$ . The usual loss is the last factor $2$ , which gives $8$ : in a second derivative the inside derivative of $2x$ appears TWICE, once per differentiation.

c) $y' = \frac{3}{2}(4 - x^2)^{1/2} \cdot (-2x) = -3x(4 - x^2)^{1/2}$ . Product rule: $y'' = -3(4 - x^2)^{1/2} + (-3x) \cdot \frac{1}{2}(4 - x^2)^{-1/2}(-2x) = -3(4 - x^2)^{1/2} + 3x^2(4 - x^2)^{-1/2}$ . Lowest power $-\frac{1}{2}$ : $y'' = (4 - x^2)^{-1/2}\left[-3(4 - x^2) + 3x^2\right] = \frac{6x^2 - 12}{\sqrt{4 - x^2}}$ . At $x = 1$ : $\frac{-6}{\sqrt 3} = -2\sqrt 3 \approx -3.4641$ , rationalised by $\frac{6}{\sqrt 3} = \frac{6\sqrt 3}{3}$ .

d) First derivative: $g'(x) = f'(x^2) \cdot 2x$ , a product. Product rule: $g''(x) = \left[f'(x^2)\right]' \cdot 2x + f'(x^2) \cdot 2$ , and $\left[f'(x^2)\right]' = f''(x^2) \cdot 2x$ by the chain rule once more. So $g''(x) = 4x^2 f''(x^2) + 2f'(x^2)$ . At $x = 2$ : $x^2 = 4$ , so $g''(2) = 16 \cdot 3 + 2 \cdot (-1) = 46$ . The student's $48 = 16 \cdot 3$ is the first term alone: the product rule was skipped and the derivative of the factor $2x$ forgotten. Note that $f'$ and $f''$ are read at $x^2 = 4$ , never at $2$ .

e) $y' = -1 \cdot (1 - 2x)^{-2} \cdot (-2) = 2(1 - 2x)^{-2}$ . $y'' = 2 \cdot (-2)(1 - 2x)^{-3} \cdot (-2) = 8(1 - 2x)^{-3}$ . $y''' = 8 \cdot (-3)(1 - 2x)^{-4} \cdot (-2) = 48(1 - 2x)^{-4}$ . Each differentiation brings down the exponent, $-(k + 1)$ , and the inside derivative $-2$ : their product is $2(k + 1)$ , positive. Hence $y^{(n)} = 2^n\, n!\,(1 - 2x)^{-(n+1)}$ , which matches $2, 8, 48$ . At $x = 0$ : $y'''(0) = 48$ and $y^{(5)}(0) = 2^5 \cdot 120 = 3840$ . Forgetting the inside factor gives $n!$ with alternating signs, the pattern of $\frac{1}{1 + x}$ instead.

Exercise 7: Tangent lines to a logistic curve

The figure shows the curve $y = \frac{4}{1 + e^{-x}}$ , which rises from $0$ toward the dashed line $y = 4$ . It is a composite built in layers: a reciprocal, then $1 + w$ , then $w = e^{-x}$ , itself an exponential of $-x$ .

The tangent line at $x = a$ is $y = y(a) + y'(a)(x - a)$ . On this curve, the derivative is easy to get and awkward to use until it is REWRITTEN with the laws of exponents; that rewriting is the heart of the exercise.

a) Write $y$ as a power, differentiate, and show that $y' = \frac{4e^{-x}}{(1 + e^{-x})^2} = \frac{4e^{x}}{(e^{x} + 1)^2}$ . Give $y'(\ln 2)$ .
b) Find the tangent line at $x = 0$ . Explain why the curve has no horizontal tangent.
c) Find the points of the curve where the tangent has slope $\frac{3}{4}$ . (Hint: set $w = e^{-x}$ .)
d) Give the equations of the two tangent lines of c), with their $y$ -intercepts to four decimals.
e) Show that $y(x) + y(-x) = 4$ for all $x$ . Differentiate this identity with the chain rule to compare $y'(-x)$ and $y'(x)$ , and explain the two points of c).

Show the solution

Answers

a) $y' = \frac{4e^{-x}}{(1 + e^{-x})^2} = \frac{4e^{x}}{(e^{x} + 1)^2}$ , $y'(\ln 2) = \frac{8}{9}$
b) $y = 2 + x$ ; $y' > 0$ everywhere, so no horizontal tangent.
c) $(\ln 3, 3)$ and $(-\ln 3, 1)$
d) $y = \frac{3}{4}x + 3 - \frac{3}{4}\ln 3$ (intercept $\approx 2.1760$ ) and $y = \frac{3}{4}x + 1 + \frac{3}{4}\ln 3$ (intercept $\approx 1.8240$ )
e) Differentiating gives $y'(x) - y'(-x) = 0$ : $y'(-x) = y'(x)$ , so the points of c) are symmetric about $(0, 2)$ .

a) $y = 4(1 + e^{-x})^{-1}$ . Outside: $4u^{-1}$ , derivative $-4u^{-2}$ ; inside: $u = 1 + e^{-x}$ , and $u' = e^{-x} \cdot (-1) = -e^{-x}$ , itself a chain rule with inside $-x$ . So $y' = -4(1 + e^{-x})^{-2} \cdot (-e^{-x}) = \frac{4e^{-x}}{(1 + e^{-x})^2}$ . Two minus signs cancel, and the result is positive, as the rising curve requires. Multiply top and bottom by $e^{2x}$ : $4e^{-x}e^{2x} = 4e^{x}$ and $(1 + e^{-x})^2 e^{2x} = \left(e^{x}(1 + e^{-x})\right)^2 = (e^{x} + 1)^2$ , so $y' = \frac{4e^{x}}{(e^{x} + 1)^2}$ . At $x = \ln 2$ : $e^{\ln 2} = 2$ , so $y'(\ln 2) = \frac{8}{9}$ .

b) At $x = 0$ : $y(0) = \frac{4}{2} = 2$ and $y'(0) = \frac{4}{4} = 1$ , so the tangent is $y = 2 + x$ . For every $x$ , $e^{x} > 0$ and $(e^{x} + 1)^2 > 0$ , so $y' > 0$ : the numerator is never zero and the curve has no horizontal tangent. Its slope tends to $0$ at both ends without ever reaching it.

c) Solve $\frac{4e^{-x}}{(1 + e^{-x})^2} = \frac{3}{4}$ . With $w = e^{-x} > 0$ : $16w = 3(1 + w)^2$ , so $3w^2 - 10w + 3 = 0$ , that is $(3w - 1)(w - 3) = 0$ . Then $w = 3$ gives $e^{-x} = 3$ , $x = -\ln 3$ , and $y = \frac{4}{1 + 3} = 1$ ; $w = \frac{1}{3}$ gives $x = \ln 3$ and $y = \frac{4}{1 + 1/3} = 3$ . The points are $(\ln 3, 3)$ and $(-\ln 3, 1)$ . The substitution turns an exponential equation into a quadratic, and both roots are kept because both are positive.

d) At $(\ln 3, 3)$ : $y = 3 + \frac{3}{4}(x - \ln 3) = \frac{3}{4}x + 3 - \frac{3}{4}\ln 3$ , intercept $\approx 2.1760$ . At $(-\ln 3, 1)$ : $y = 1 + \frac{3}{4}(x + \ln 3) = \frac{3}{4}x + 1 + \frac{3}{4}\ln 3$ , intercept $\approx 1.8240$ . The two lines are parallel, and their intercepts add up to $4$ : their midpoint is on the line $y = 2$ , which part e) explains. The solution figure shows the three tangents.

e) $y(-x) = \frac{4}{1 + e^{x}}$ and $y(x) = \frac{4}{1 + e^{-x}} = \frac{4e^{x}}{e^{x} + 1}$ , so $y(x) + y(-x) = \frac{4e^{x} + 4}{e^{x} + 1} = 4$ . Differentiate both sides. The term $y(-x)$ is a composite with inside function $-x$ , of derivative $-1$ : $\frac{d}{dx}y(-x) = -y'(-x)$ . So $y'(x) - y'(-x) = 0$ , that is $y'(-x) = y'(x)$ . The curve is symmetric about the point $(0, 2)$ , and symmetric points carry equal slopes: $x = \ln 3$ and $x = -\ln 3$ both have slope $\frac{3}{4}$ , with heights $3$ and $1$ symmetric about $2$ . Forgetting the factor $-1$ from the inside function would give $y'(-x) = -y'(x)$ , impossible for a curve whose slope is always positive.

Exercise 8: Five statements to correct

Each statement below was written by a student on a MATH 203 assignment, and each is false. Say what is wrong, give a counterexample with numbers, and write the correct statement. Three of the five errors are errors of algebra, not of calculus.

a) $\frac{d}{dx}(3x - 1)^{-1/2} = -\frac{3}{2}(3x - 1)^{1/2}$ .
b) The derivative of a square is the square of the derivative, so $\frac{d}{dx}\cos^2 x = (-\sin x)^2 = \sin^2 x$ .
c) $\frac{d}{dx}\left(e^{x}\right)^3 = 3e^{2x}$ .
d) The derivative of $x^3\sqrt{2x + 1}$ simplifies to $x^2\sqrt{2x + 1}\,(x + 3)$ .
e) If $h = f \circ g$ and $h'(a) = 0$ , then $g'(a) = 0$ .

Type your answers, the page tells you right or wrong 0/8

a)

The statement is

Correct derivative at

x = 1

b)

The statement is

Correct slope at

x = \frac{\pi}{4}

c)

The statement is

d)

The statement is

Correct derivative at

x = 4

e)

The statement is

Show the solution

Answers

a) False: the exponent is $-\frac{3}{2}$ ; the derivative is $-\frac{3}{2}(3x - 1)^{-3/2}$ , $\approx -0.5303$ at $x = 1$ .
b) False: $\frac{d}{dx}\cos^2 x = -2\sin x\cos x$ ; at $x = \frac{\pi}{4}$ , $-1$ and not $\frac{1}{2}$ .
c) False: $(e^x)^3 = e^{3x}$ , derivative $3e^{3x}$ ; at $x = 1$ , $3e^3$ and not $3e^2$ .
d) False: the derivative is $\frac{x^2(7x + 3)}{\sqrt{2x + 1}}$ ; at $x = 4$ , $\frac{496}{3}$ and not $336$ .
e) False: $h'(a) = f'(g(a))\,g'(a)$ can vanish through $f'(g(a))$ ; $f(u) = u^2$ , $g(x) = x - 1$ , $a = 1$ .

a) FALSE. The inside factor $3$ is there, but the new exponent is wrong: $-\frac{1}{2} - 1 = -\frac{3}{2}$ , not $+\frac{1}{2}$ . The student subtracted $1$ in the wrong direction, a slip of arithmetic on fractions. Correct: $\frac{d}{dx}(3x - 1)^{-1/2} = -\frac{1}{2}(3x - 1)^{-3/2} \cdot 3 = -\frac{3}{2(3x - 1)^{3/2}}$ . At $x = 1$ : $-\frac{3}{2} \cdot 2^{-3/2} \approx -0.5303$ , while the claim gives $-\frac{3}{2}\sqrt 2 \approx -2.1213$ . Check of size: the function $\frac{1}{\sqrt{3x - 1}}$ flattens as $x$ grows, so its slope must shrink toward $0$ , never grow like a square root.

b) FALSE. $\cos^2 x = (\cos x)^2$ is a composite: the outside function is the square, the inside is $u = \cos x$ . The outside derivative is $2u$ , evaluated at the inside left alone, then multiplied by $u' = -\sin x$ : $\frac{d}{dx}\cos^2 x = 2\cos x \cdot (-\sin x) = -2\sin x\cos x = -\sin 2x$ . Counterexample at $x = \frac{\pi}{4}$ : the true slope is $-2 \cdot \frac{\sqrt 2}{2} \cdot \frac{\sqrt 2}{2} = -1$ , while the claim gives $\frac{1}{2}$ , and even a POSITIVE slope, for a function that decreases on $\left(0, \frac{\pi}{2}\right)$ . Squaring the derivative of the inside is not a rule; the square is differentiated as a square, $2u$ , and the inside derivative multiplies.

c) FALSE. By the laws of exponents, $(e^x)^3 = e^{3x}$ , whose derivative is $3e^{3x}$ (inside function $3x$ ). The student applied the power rule to the cube and stopped: $3(e^x)^2 = 3e^{2x}$ , without the inside derivative $e^x$ . Done correctly, the Power Chain Rule gives $3(e^x)^2 \cdot e^x = 3e^{3x}$ , the same answer. At $x = 1$ : $3e^3 \approx 60.26$ against $3e^2 \approx 22.17$ . At $x = 0$ both give $3$ , which is why a check at $0$ would not catch it.

d) FALSE. By the product and chain rules, $\frac{d}{dx}x^3(2x + 1)^{1/2} = 3x^2(2x + 1)^{1/2} + x^3 \cdot \frac{1}{2}(2x + 1)^{-1/2} \cdot 2 = 3x^2(2x + 1)^{1/2} + x^3(2x + 1)^{-1/2}$ . The student factored out $(2x + 1)^{1/2}$ , the HIGHER power, and left $x^3$ in the second term, when it should have left $x^3(2x + 1)^{-1}$ . The lowest power is $-\frac{1}{2}$ : $x^2(2x + 1)^{-1/2}\left[3(2x + 1) + x\right] = \frac{x^2(7x + 3)}{\sqrt{2x + 1}}$ . At $x = 4$ : the unfactored sum is $144 + \frac{64}{3} = \frac{496}{3}$ , the correct form gives $\frac{16 \cdot 31}{3} = \frac{496}{3}$ , and the claim gives $16 \cdot 3 \cdot 7 = 336$ . Evaluating both forms at one convenient point is the cheapest check of any factorisation.

e) FALSE. By the chain rule, $h'(a) = f'(g(a)) \cdot g'(a)$ , a product: it vanishes when EITHER factor does. Take $f(u) = u^2$ , $g(x) = x - 1$ and $a = 1$ : $h(x) = (x - 1)^2$ and $h'(1) = 0$ , yet $g'(1) = 1$ . Here the zero comes from $f'(g(1)) = f'(0) = 0$ . Correct statement: for differentiable $f$ and $g$ , $h'(a) = 0$ if and only if $f'(g(a)) = 0$ or $g'(a) = 0$ . This is why the horizontal tangents of a composite come from the zeros of the outside factor AND of the inside factor.

Exercise 9: A radiosonde in cascade: degrees per kilometre, kilometres per minute

A weather balloon carrying a radiosonde is released at $t = 0$ and rises at $5$ m/s, so its altitude after $t$ minutes is $h = 0.3t$ kilometres ( $0 \le t \le 100$ ). The air temperature $T$ , in degrees Celsius, depends on the altitude as in the International Standard Atmosphere, drawn in the figure: $T = 15 - 6.5h$ for $0 \le h \le 11$ , $T = -56.5$ for $11 \le h \le 20$ , and $T = -56.5 + (h - 20)$ for $20 \le h \le 32$ . The air pressure is modelled by $P = 100e^{-h/7}$ kilopascals.

The temperature the radiosonde reads depends on the altitude, which depends on the time: $\frac{dT}{dt} = \frac{dT}{dh} \cdot \frac{dh}{dt}$ , each derivative read at the current value of its OWN variable, and the units cancel like fractions. The calculator is used for the final decimals only.

a) Give the units of $\frac{dT}{dh}$ , $\frac{dh}{dt}$ and $\frac{dT}{dt}$ , then compute $\frac{dT}{dt}$ at $t = 20$ minutes, with a sentence of interpretation.
b) Compute $\frac{dT}{dt}$ at $t = 50$ and at $t = 80$ minutes. What does the sign at $t = 80$ say?
c) At what time does the balloon reach $11$ km? Does $\frac{dT}{dt}$ exist at that instant? Give the one-sided rates.
d) Find $\frac{dP}{dh}$ , then $\frac{dP}{dt}$ at $t = 35$ minutes, in kPa per minute to four decimals.
e) In a simplified gas model the balloon's volume is $V = \frac{T + 273}{P}$ cubic metres ( $2.88$ m³ at launch). Rewrite $V$ as a function of $h$ without a fraction in $e$ , find $\frac{dV}{dh}$ in factored form, and $\frac{dV}{dt}$ at $t = 20$ minutes.

Show the solution

Answers

a) °C per km, km per min, °C per min; $\frac{dT}{dt} = -6.5 \cdot 0.3 = -1.95$ °C/min at $t = 20$
b) $0$ °C/min at $t = 50$ ; $+0.3$ °C/min at $t = 80$ (the air warms above $20$ km)
c) $t = \frac{110}{3} \approx 36.67$ min; $\frac{dT}{dt}$ does not exist: $-1.95$ on the left, $0$ on the right
d) $\frac{dP}{dh} = -\frac{100}{7}e^{-h/7}$ ; $\frac{dP}{dt} = -\frac{30}{7}e^{-1.5} \approx -0.9563$ kPa/min
e) $V = \frac{(T + 273)e^{h/7}}{100}$ , $\frac{dV}{dh} = \frac{e^{h/7}}{100}\left[T'(h) + \frac{T + 273}{7}\right]$ ; $\frac{dV}{dt} \approx 0.2055$ m³/min

a) $\frac{dT}{dh}$ is a change of degrees per change of kilometres: °C per km. $\frac{dh}{dt}$ : km per minute. Their product: $\frac{\text{°C}}{\text{km}} \cdot \frac{\text{km}}{\text{min}} = \frac{\text{°C}}{\text{min}}$ , the unit of $\frac{dT}{dt}$ . At $t = 20$ : $h = 6$ km, in the first piece, where $\frac{dT}{dh} = -6.5$ ; and $\frac{dh}{dt} = 0.3$ . So $\frac{dT}{dt} = -6.5 \cdot 0.3 = -1.95$ °C/min: twenty minutes after launch, the radiosonde reads a temperature falling by about $2$ degrees each minute. The slope of $T$ is read at $h = 6$ , the current altitude, not at $20$ , which is a number of minutes.

b) At $t = 50$ : $h = 15$ km, on the flat piece, so $\frac{dT}{dh} = 0$ and $\frac{dT}{dt} = 0 \cdot 0.3 = 0$ °C/min, whatever the speed of the balloon. At $t = 80$ : $h = 24$ km, where $\frac{dT}{dh} = 1$ , so $\frac{dT}{dt} = 1 \cdot 0.3 = 0.3$ °C/min. The positive sign says the radiosonde reads a WARMING air while still climbing: above $20$ km the ozone layer heats the stratosphere. A product of rates changes sign when one link does.

c) $0.3t = 11$ gives $t = \frac{110}{3} \approx 36.67$ min. At $h = 11$ the graph of $T$ has a corner: slope $-6.5$ on the left, $0$ on the right, so $T$ is not differentiable at $h(t)$ , and the hypothesis of the chain rule fails. Look at each side: just before, $\frac{dT}{dt} = -6.5 \cdot 0.3 = -1.95$ °C/min; just after, $0$ . The one-sided rates differ, so $\frac{dT}{dt}$ does not exist at that instant. Checking that the outside function is differentiable at the INSIDE value is part of applying the chain rule.

d) $P = 100e^{u}$ with inside $u = -\frac{h}{7}$ and $u' = -\frac{1}{7}$ , so $\frac{dP}{dh} = -\frac{100}{7}e^{-h/7}$ kPa per km, which is also $-\frac{P}{7}$ . At $t = 35$ : $h = 10.5$ , $e^{-10.5/7} = e^{-1.5}$ , and $\frac{dP}{dt} = \frac{dP}{dh} \cdot \frac{dh}{dt} = -\frac{100}{7}e^{-1.5} \cdot 0.3 = -\frac{30}{7}e^{-1.5} \approx -0.9563$ kPa/min. Units: $\frac{\text{kPa}}{\text{km}} \cdot \frac{\text{km}}{\text{min}} = \frac{\text{kPa}}{\text{min}}$ .

e) Dividing by $P = 100e^{-h/7}$ is multiplying by $\frac{e^{h/7}}{100}$ , a law of exponents: $V = \frac{1}{100}(T + 273)e^{h/7}$ , a product with no fraction left. Product rule, chain rule on $e^{h/7}$ (inside $\frac{h}{7}$ , derivative $\frac{1}{7}$ ): $\frac{dV}{dh} = \frac{1}{100}\left[T'(h)e^{h/7} + (T + 273)\frac{1}{7}e^{h/7}\right] = \frac{e^{h/7}}{100}\left[T'(h) + \frac{T + 273}{7}\right]$ , the common factor $e^{h/7}$ taken out. At $t = 20$ : $h = 6$ , $T = 15 - 39 = -24$ , $T + 273 = 249$ , $T'(6) = -6.5$ , so the bracket is $\frac{249}{7} - 6.5 = \frac{203.5}{7} \approx 29.07$ and $\frac{dV}{dh} = \frac{e^{6/7}}{100} \cdot \frac{203.5}{7} \approx 0.6850$ m³ per km. Then $\frac{dV}{dt} = 0.6850 \cdot 0.3 \approx 0.2055$ m³/min. The cooling alone would shrink the balloon; the bracket shows that the falling pressure wins, by $29.07$ against $6.5$ .

Exercise 10: Wind chill during a storm: a cascade with a real exponent

The wind chill index used by Environment and Climate Change Canada is $W = 13.12 + 0.6215T - 11.37V^{0.16} + 0.3965\,T\,V^{0.16}$ , where $T$ is the air temperature in °C and $V$ the wind speed in km/h; $W$ is read like a temperature, in °C. On a day when the air stays at $T = -20$ °C, a storm sets in: $t$ hours after noon, the wind blows at $V = 20 + 6t$ km/h ( $0 \le t \le 5$ ).

The wind chill depends on the wind, which depends on the time. The exponent $0.16$ is a real number: the Power Rule still applies, and the exponent of the derivative is $0.16 - 1 = -0.84$ , a NEGATIVE exponent that moves $V$ to the denominator. The figure shows $W$ as a function of $V$ at $T = -20$ °C.

a) Show that at $T = -20$ °C, $W = 0.69 - 19.3V^{0.16}$ . Compute $W$ at $V = 32$ km/h to two decimals.
b) Find $\frac{dW}{dV}$ and its value at $V = 32$ km/h, with its unit, to four decimals.
c) Find $\frac{dW}{dt}$ at $t = 2$ hours, in °C per hour to two decimals. A student computes $\frac{dW}{dV}$ at $V = 2$ : what does that give, and why is it absurd?
d) Write $W$ directly as a function of $t$ , differentiate with the chain rule, and check c).
e) Compute $\frac{dW}{dt}$ at $t = 0$ and at $t = 5$ . The wind rises at a constant rate, yet the wind chill drops more and more slowly: which link of the chain explains it?

Show the solution

Answers

a) $13.12 - 12.43 = 0.69$ and $-11.37 - 7.93 = -19.3$ ; $W(32) = 0.69 - 19.3 \cdot 2^{0.8} \approx -32.91$ °C
b) $\frac{dW}{dV} = -3.088V^{-0.84} = -\frac{3.088}{V^{0.84}}$ ; $\approx -0.1680$ °C per km/h at $V = 32$
c) $\frac{dW}{dt} \approx -0.1680 \cdot 6 \approx -1.01$ °C/h; $\frac{dW}{dV}$ at $V = 2$ reads the wind at a number of hours.
d) $W(t) = 0.69 - 19.3(20 + 6t)^{0.16}$ , $W'(t) = -18.528(20 + 6t)^{-0.84}$ , $W'(2) \approx -1.01$ °C/h
e) $\approx -1.50$ °C/h at $t = 0$ , $\approx -0.69$ °C/h at $t = 5$ ; the link $\frac{dW}{dV}$ shrinks as $V$ grows.

a) At $T = -20$ : $0.6215 \cdot (-20) = -12.43$ , so the constant part is $13.12 - 12.43 = 0.69$ ; and $0.3965 \cdot (-20) = -7.93$ , so the coefficient of $V^{0.16}$ is $-11.37 - 7.93 = -19.3$ . Hence $W = 0.69 - 19.3V^{0.16}$ . At $V = 32 = 2^5$ : $32^{0.16} = 2^{0.8} \approx 1.7411$ , so $W \approx 0.69 - 33.60 = -32.91$ °C. A power of a power multiplies exponents, $5 \cdot 0.16 = 0.8$ : the calculator is not even needed until the last step.

b) Power Rule: $\frac{dW}{dV} = -19.3 \cdot 0.16\,V^{0.16 - 1} = -3.088V^{-0.84} = -\frac{3.088}{V^{0.84}}$ . The exponent $0.16 - 1 = -0.84$ is where students lose the mark, writing $1.16$ or $-1.16$ . At $V = 32$ : $32^{-0.84} = 2^{-4.2} \approx 0.05441$ , so $\frac{dW}{dV} \approx -0.1680$ °C per km/h: near $32$ km/h, each extra km/h of wind makes it feel about $0.17$ degree colder.

c) Chain rule: $\frac{dW}{dt} = \frac{dW}{dV} \cdot \frac{dV}{dt}$ . At $t = 2$ : $V = 32$ km/h, $\frac{dW}{dV} \approx -0.1680$ , and $\frac{dV}{dt} = 6$ km/h per hour. So $\frac{dW}{dt} \approx -1.008$ , about $-1.01$ °C per hour. Units: $\frac{\text{°C}}{\text{km/h}} \cdot \frac{\text{km/h}}{\text{h}} = \frac{\text{°C}}{\text{h}}$ . The student who evaluates $\frac{dW}{dV}$ at $V = 2$ gets $-3.088 \cdot 2^{-0.84} \approx -1.73$ , then $-10.35$ °C/h after multiplying by $6$ : a wind chill falling ten degrees an hour. The $2$ is a number of HOURS fed to a function whose input is a wind speed; the outside derivative must be read at the inside VALUE, $V(2) = 32$ .

d) $W(t) = 0.69 - 19.3(20 + 6t)^{0.16}$ . With the inside $u = 20 + 6t$ and $u' = 6$ : $W'(t) = -19.3 \cdot 0.16(20 + 6t)^{-0.84} \cdot 6 = -18.528(20 + 6t)^{-0.84}$ . At $t = 2$ : $-18.528 \cdot 32^{-0.84} \approx -1.008$ °C/h, as in c). Composing first and differentiating once, or differentiating link by link: both roads must meet, and checking that they do is the cheapest verification there is.

e) At $t = 0$ : $V = 20$ , $W'(0) = -18.528 \cdot 20^{-0.84} \approx -1.50$ °C/h. At $t = 5$ : $V = 50$ , $W'(5) = -18.528 \cdot 50^{-0.84} \approx -0.69$ °C/h. The link $\frac{dV}{dt} = 6$ is constant; the link $\frac{dW}{dV} = -\frac{3.088}{V^{0.84}}$ shrinks in size as $V$ grows, because of the negative exponent. The figure shows it: the curve is steep at low wind and flattens. The first few km/h of wind cost the most degrees, which is why a light breeze on a cold day feels so much worse than calm air.

Corrected exercises: the chain rule (MATH 203)

Course recap

Part A: the basics (/50)

Exercise 1: Powers of a function: rewrite the exponent, then name u

Exercise 2: Three and four layers: one factor per layer

Exercise 3: The chain rule from a table of values

Exercise 4: Exponentials in base a: the laws of exponents first

Exercise 5: Factor the lowest power: the algebra after the chain rule

Part B: problems and reasoning (/50)

Exercise 6: Second and higher derivatives: the chain inside the product

Exercise 7: Tangent lines to a logistic curve

Exercise 8: Five statements to correct

Exercise 9: A radiosonde in cascade: degrees per kilometre, kilometres per minute

Exercise 10: Wind chill during a storm: a cascade with a real exponent

See also

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