Exercise 1: The six derivatives, and the identity that comes BEFORE differentiating

The six derivatives are $(\sin x)' = \cos x$ , $(\cos x)' = -\sin x$ , $(\tan x)' = \sec^2 x$ , $(\cot x)' = -\csc^2 x$ , $(\sec x)' = \sec x\tan x$ and $(\csc x)' = -\csc x\cot x$ , with $x$ in radians. They are used here with the sum and constant multiple rules only.

Four of the five functions below are much simpler than they look. Rewriting them in sines and cosines, bringing a complex fraction to a single fraction and using $\sin^2 x + \cos^2 x = 1$ turns a page of quotient rule into one line. The algebra is where the marks are: cancel only a FACTOR, never a term, and say where the simplification is valid.

a) Differentiate $f(x) = 7\sin x - 4\csc x + 3x^2$ and find the exact value of $f'\left(\frac{\pi}{6}\right)$ .
b) Simplify $g(x) = \frac{\sec x - \cos x}{\sin x}$ to a single trigonometric function, then find $g'(x)$ and $g'\left(\frac{\pi}{3}\right)$ .
c) Simplify $h(x) = \frac{\tan^2 x}{\sec x + 1}$ , stating where your simplification holds, then find $h'(x)$ and $h'\left(\frac{\pi}{3}\right)$ .
d) Let $k(x) = \tan x + \cot x$ . Differentiate term by term, write $k'(x)$ as a single fraction in $\sin x$ and $\cos x$ , and find $k'\left(\frac{\pi}{4}\right)$ and $k'\left(\frac{\pi}{6}\right)$ .
e) Let $m(x) = \frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}$ . Bring $m$ to a single fraction, show that $m(x) = 2\sec x$ , then find $m'\left(\frac{\pi}{4}\right)$ .

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Answers

a) $f'(x) = 7\cos x + 4\csc x\cot x + 6x$ ; $f'\left(\frac{\pi}{6}\right) = \frac{23\sqrt 3}{2} + \pi$
b) $g(x) = \tan x$ ; $g'(x) = \sec^2 x$ ; $g'\left(\frac{\pi}{3}\right) = 4$
c) $h(x) = \sec x - 1$ where $\cos x \ne 0$ and $\cos x \ne -1$ ; $h'(x) = \sec x\tan x$ ; $h'\left(\frac{\pi}{3}\right) = 2\sqrt 3$
d) $k'(x) = \sec^2 x - \csc^2 x = \frac{\sin^2 x - \cos^2 x}{\sin^2 x\cos^2 x}$ ; $k'\left(\frac{\pi}{4}\right) = 0$ , $k'\left(\frac{\pi}{6}\right) = -\frac{8}{3}$
e) $m(x) = 2\sec x$ ; $m'(x) = 2\sec x\tan x$ ; $m'\left(\frac{\pi}{4}\right) = 2\sqrt 2$

a) Sum and constant multiple rules, term by term: $f'(x) = 7\cos x - 4(-\csc x\cot x) + 6x = 7\cos x + 4\csc x\cot x + 6x$ . The minus written in front of $\csc x$ meets the minus of $(\csc x)'$ , and the two make a plus: this is the sign slip of the chapter. At $\frac{\pi}{6}$ : $\cos\frac{\pi}{6} = \frac{\sqrt 3}{2}$ , $\csc\frac{\pi}{6} = \frac{1}{\sin(\pi/6)} = 2$ , $\cot\frac{\pi}{6} = \frac{\cos(\pi/6)}{\sin(\pi/6)} = \sqrt 3$ . So $f'\left(\frac{\pi}{6}\right) = \frac{7\sqrt 3}{2} + 4 \cdot 2\sqrt 3 + 6\cdot\frac{\pi}{6} = \frac{7\sqrt 3}{2} + \frac{16\sqrt 3}{2} + \pi = \frac{23\sqrt 3}{2} + \pi$ , about $23.0602$ .

b) Rewrite the numerator in cosines and bring it to one fraction: $\sec x - \cos x = \frac{1}{\cos x} - \frac{\cos^2 x}{\cos x} = \frac{1 - \cos^2 x}{\cos x} = \frac{\sin^2 x}{\cos x}$ , by the Pythagorean identity. Dividing by $\sin x$ is multiplying by $\frac{1}{\sin x}$ : $g(x) = \frac{\sin^2 x}{\cos x\sin x} = \frac{\sin x}{\cos x} = \tan x$ , wherever $\sin x \ne 0$ and $\cos x \ne 0$ . Hence $g'(x) = \sec^2 x$ and $g'\left(\frac{\pi}{3}\right) = \frac{1}{\cos^2(\pi/3)} = \frac{1}{1/4} = 4$ . The quotient rule applied to the original form also works, but it takes half a page and three chances to lose a sign. The formula $\sec^2 x$ is valid only on the domain of $g$ : at $x = \pi$ , where $g$ does not exist, neither does $g'$ .

c) Use the disguise $\tan^2 x = \sec^2 x - 1$ : the numerator becomes $\sec^2 x - 1 = (\sec x - 1)(\sec x + 1)$ , a difference of squares. The factor $\sec x + 1$ cancels with the denominator, which is legitimate wherever it is not $0$ , that is $\cos x \ne -1$ ; and $\cos x \ne 0$ for $\tan x$ and $\sec x$ to exist. There $h(x) = \sec x - 1$ , so $h'(x) = \sec x\tan x$ and $h'\left(\frac{\pi}{3}\right) = 2\cdot\sqrt 3 = 2\sqrt 3$ , about $3.4641$ . The gesture is to see $\tan^2 x$ as $\sec^2 x - 1$ BECAUSE the denominator is written with $\sec x$ : the identity is chosen by the rest of the fraction. The quotient rule on the original form works too, but it needs $(\tan^2 x)'$ , a product, and a page of simplification.

d) Term by term: $k'(x) = \sec^2 x - \csc^2 x = \frac{1}{\cos^2 x} - \frac{1}{\sin^2 x}$ . Common denominator $\sin^2 x\cos^2 x$ : $k'(x) = \frac{\sin^2 x - \cos^2 x}{\sin^2 x\cos^2 x}$ . At $\frac{\pi}{4}$ , $\sin^2 x = \cos^2 x = \frac{1}{2}$ , so $k'\left(\frac{\pi}{4}\right) = 0$ : the graph of $k$ has a horizontal tangent there. At $\frac{\pi}{6}$ , $\sin^2 x = \frac{1}{4}$ and $\cos^2 x = \frac{3}{4}$ : $k'\left(\frac{\pi}{6}\right) = \frac{1/4 - 3/4}{3/16} = -\frac{1}{2}\cdot\frac{16}{3} = -\frac{8}{3}$ . Check with the first form: $\sec^2\frac{\pi}{6} = \frac{4}{3}$ and $\csc^2\frac{\pi}{6} = 4$ , and $\frac{4}{3} - 4 = -\frac{8}{3}$ . The minus of $(\cot x)'$ is what produces a difference and not a sum.

e) Common denominator $(1 + \sin x)\cos x$ : the numerator is $\cos^2 x + (1 + \sin x)^2 = \cos^2 x + 1 + 2\sin x + \sin^2 x$ . Collect $\sin^2 x + \cos^2 x = 1$ : the numerator is $2 + 2\sin x = 2(1 + \sin x)$ . Cancel the FACTOR $1 + \sin x$ , which is not zero where $m$ is defined (if $\sin x = -1$ then $\cos x = 0$ ): $m(x) = \frac{2(1 + \sin x)}{(1 + \sin x)\cos x} = \frac{2}{\cos x} = 2\sec x$ . Hence $m'(x) = 2\sec x\tan x$ and $m'\left(\frac{\pi}{4}\right) = 2\cdot\sqrt 2\cdot 1 = 2\sqrt 2$ , about $2.8284$ . Three algebra gestures in a row, expand the square, spot the identity, factor the $2$ : this is the exact place where a MATH 203 derivative is won or lost.

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Exercise 2: The product rule with trigonometric factors: name u and v, then collect

The product rule $(uv)' = u'v + uv'$ does not change when a factor is trigonometric; what changes is the number of places where a sign can be lost. Write $u$ , $v$ , $u'$ and $v'$ on your copy before assembling anything. For three factors, $(uvw)' = u'vw + uv'w + uvw'$ .

A derivative is not finished when the rule has been applied: expand, COLLECT like terms and factor, because the exam expects the simplest form and because terms often cancel.

a) Differentiate $y = x^3\cos x$ , factor the result, and find $y'(\pi)$ .
b) Differentiate $y = \sqrt{x}\,\tan x$ , write $y'$ as a single fraction, and find $y'\left(\frac{\pi}{4}\right)$ .
c) Differentiate $y = (x^2 - 2)\sin x + 2x\cos x$ and simplify until one term is left. Find $y'\left(\frac{\pi}{3}\right)$ .
d) Differentiate $y = x e^x\sin x$ (three factors) and find $y'(\pi)$ .
e) Differentiate $y = \csc x\cot x$ and use $\cot^2 x = \csc^2 x - 1$ to write $y'$ in terms of $\csc x$ only. Find $y'\left(\frac{\pi}{6}\right)$ .

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Answers

a) $y' = 3x^2\cos x - x^3\sin x = x^2(3\cos x - x\sin x)$ ; $y'(\pi) = -3\pi^2$
b) $y' = \frac{\tan x + 2x\sec^2 x}{2\sqrt x}$ ; $y'\left(\frac{\pi}{4}\right) = \frac{1 + \pi}{\sqrt\pi}$
c) $y' = x^2\cos x$ ; $y'\left(\frac{\pi}{3}\right) = \frac{\pi^2}{18}$
d) $y' = e^x\left[(1 + x)\sin x + x\cos x\right]$ ; $y'(\pi) = -\pi e^{\pi}$
e) $y' = -\csc x\left(\cot^2 x + \csc^2 x\right) = -\csc x\left(2\csc^2 x - 1\right)$ ; $y'\left(\frac{\pi}{6}\right) = -14$

a) $u = x^3$ , $u' = 3x^2$ , $v = \cos x$ , $v' = -\sin x$ . So $y' = 3x^2\cos x + x^3(-\sin x) = 3x^2\cos x - x^3\sin x = x^2(3\cos x - x\sin x)$ , the smallest power $x^2$ factored out. At $\pi$ : $\cos\pi = -1$ and $\sin\pi = 0$ , so $y'(\pi) = \pi^2(-3 - 0) = -3\pi^2$ , about $-29.6088$ . Writing $(x^3\cos x)' = 3x^2(-\sin x)$ , the product of the derivatives, is worth nothing.

b) Rewrite the root as a power first: $\sqrt x = x^{1/2}$ , so $u' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt x}$ . With $v = \tan x$ and $v' = \sec^2 x$ : $y' = \frac{\tan x}{2\sqrt x} + \sqrt x\sec^2 x$ . Single fraction over $2\sqrt x$ : $\sqrt x\sec^2 x = \frac{2x\sec^2 x}{2\sqrt x}$ , because $\sqrt x\cdot 2\sqrt x = 2x$ . So $y' = \frac{\tan x + 2x\sec^2 x}{2\sqrt x}$ , for $x > 0$ with $\cos x \ne 0$ . At $\frac{\pi}{4}$ : $\tan = 1$ , $\sec^2 = 2$ , $2\sqrt{\pi/4} = \sqrt\pi$ , so $y'\left(\frac{\pi}{4}\right) = \frac{1 + \pi}{\sqrt\pi}$ , about $2.3366$ . The negative exponent $x^{-1/2}$ becomes a root in the DENOMINATOR, not $-\sqrt x$ .

c) First product, $u = x^2 - 2$ , $v = \sin x$ : $2x\sin x + (x^2 - 2)\cos x$ . Second product, $u = 2x$ , $v = \cos x$ : $2\cos x - 2x\sin x$ . Add and collect: the terms $2x\sin x$ and $-2x\sin x$ cancel, and $(x^2 - 2)\cos x + 2\cos x = x^2\cos x$ . So $y' = x^2\cos x$ . At $\frac{\pi}{3}$ : $y' = \frac{\pi^2}{9}\cdot\frac{1}{2} = \frac{\pi^2}{18}$ , about $0.5483$ . A student who stops at the four-term sum has not answered a question that says simplify, and misses the whole point: this function was built to have the derivative $x^2\cos x$ .

d) $u = x$ , $v = e^x$ , $w = \sin x$ , with $u' = 1$ , $v' = e^x$ , $w' = \cos x$ : $y' = e^x\sin x + xe^x\sin x + xe^x\cos x$ . Factor $e^x$ , then collect the sines: $y' = e^x\left[(1 + x)\sin x + x\cos x\right]$ . At $\pi$ : $\sin\pi = 0$ and $\cos\pi = -1$ , so $y'(\pi) = e^\pi\left[0 - \pi\right] = -\pi e^\pi$ , about $-72.6986$ . Three factors give three terms, each with exactly ONE derivative in it.

e) $u = \csc x$ , $u' = -\csc x\cot x$ , $v = \cot x$ , $v' = -\csc^2 x$ . So $y' = -\csc x\cot x\cdot\cot x + \csc x\cdot(-\csc^2 x) = -\csc x\cot^2 x - \csc^3 x = -\csc x\left(\cot^2 x + \csc^2 x\right)$ . With $\cot^2 x = \csc^2 x - 1$ , obtained by dividing $\sin^2 x + \cos^2 x = 1$ by $\sin^2 x$ : $y' = -\csc x\left(2\csc^2 x - 1\right)$ . At $\frac{\pi}{6}$ , $\csc = 2$ : $y' = -2(8 - 1) = -14$ . Check with the unsimplified form: $\cot^2\frac{\pi}{6} = 3$ , and $-2(3 + 4) = -14$ . Both derivatives are negative, so both minus signs survive the product rule.

Exercise 3: The quotient rule, then the identity that comes AFTER

The quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ has an ORDER: derivative of the top times the bottom first. Swapping the two terms changes the sign of the whole answer.

With trigonometric functions the numerator it produces almost always contains $\sin^2 x + \cos^2 x$ or $\sec^2 x - \tan^2 x$ in disguise. Expanding, then spotting the identity, is the gesture that turns an unreadable fraction into the expected answer.

a) Differentiate $q(x) = \frac{\sin x + \cos x}{\sin x - \cos x}$ and show that $q'(x) = \frac{-2}{(\sin x - \cos x)^2}$ . Find $q'(0)$ , and say whether $q$ is defined at $\frac{\pi}{4}$ .
b) Differentiate $r(x) = \frac{1 - \cos x}{1 + \cos x}$ , simplify, and find $r'\left(\frac{2\pi}{3}\right)$ .
c) Differentiate $g(x) = \frac{\tan x}{x}$ and find $g'\left(\frac{\pi}{4}\right)$ .
d) Differentiate $h(x) = \frac{\sec x}{\tan x}$ by the quotient rule and simplify with $\tan^2 x - \sec^2 x = -1$ . Check by simplifying $h$ first, and find $h'\left(\frac{\pi}{3}\right)$ .
e) Differentiate $s(x) = \frac{x^2}{\cos x}$ twice over: by the quotient rule, then as the product $x^2\sec x$ . Show that the two answers agree and find $s'\left(\frac{\pi}{3}\right)$ .

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Answers

a) $q'(x) = \frac{-2}{(\sin x - \cos x)^2}$ ; $q'(0) = -2$ ; $q$ is not defined at $\frac{\pi}{4}$ .
b) $r'(x) = \frac{2\sin x}{(1 + \cos x)^2}$ ; $r'\left(\frac{2\pi}{3}\right) = 4\sqrt 3$
c) $g'(x) = \frac{x\sec^2 x - \tan x}{x^2}$ ; $g'\left(\frac{\pi}{4}\right) = \frac{8\pi - 16}{\pi^2}$
d) $h'(x) = -\csc x\cot x$ (and $h = \csc x$ where $\cos x \ne 0$ ); $h'\left(\frac{\pi}{3}\right) = -\frac{2}{3}$
e) $s'(x) = \frac{2x\cos x + x^2\sin x}{\cos^2 x} = 2x\sec x + x^2\sec x\tan x$ ; $s'\left(\frac{\pi}{3}\right) = \frac{4\pi}{3} + \frac{2\sqrt 3\,\pi^2}{9}$

a) $u = \sin x + \cos x$ , $u' = \cos x - \sin x$ ; $v = \sin x - \cos x$ , $v' = \cos x + \sin x$ . Numerator: $u'v - uv' = (\cos x - \sin x)(\sin x - \cos x) - (\sin x + \cos x)^2 = -(\sin x - \cos x)^2 - (\sin x + \cos x)^2$ . Expand both squares: $(\sin^2 x - 2\sin x\cos x + \cos^2 x) + (\sin^2 x + 2\sin x\cos x + \cos^2 x) = 2(\sin^2 x + \cos^2 x) = 2$ . So the numerator is $-2$ and $q'(x) = \frac{-2}{(\sin x - \cos x)^2}$ . Then $q'(0) = \frac{-2}{(0 - 1)^2} = -2$ . At $\frac{\pi}{4}$ , $\sin x = \cos x$ : the denominator of $q$ vanishes, $q$ is not defined there and neither is $q'$ . The cross terms $\pm 2\sin x\cos x$ cancel: expanding is what makes the identity visible.

b) $u = 1 - \cos x$ , $u' = \sin x$ ; $v = 1 + \cos x$ , $v' = -\sin x$ . Numerator: $\sin x(1 + \cos x) - (1 - \cos x)(-\sin x) = \sin x + \sin x\cos x + \sin x - \sin x\cos x = 2\sin x$ . So $r'(x) = \frac{2\sin x}{(1 + \cos x)^2}$ , for $\cos x \ne -1$ . At $\frac{2\pi}{3}$ : $\sin = \frac{\sqrt 3}{2}$ , $1 + \cos = \frac{1}{2}$ , so $r' = \frac{\sqrt 3}{1/4} = 4\sqrt 3$ , about $6.9282$ . The minus of the rule meets the minus of $v'$ : $-(1 - \cos x)(-\sin x)$ is a PLUS. Lose it and the $\sin x\cos x$ terms add instead of cancelling.

c) $u = \tan x$ , $u' = \sec^2 x$ , $v = x$ , $v' = 1$ : $g'(x) = \frac{x\sec^2 x - \tan x}{x^2}$ , for $x \ne 0$ and $\cos x \ne 0$ . At $\frac{\pi}{4}$ : $\frac{\frac{\pi}{4}\cdot 2 - 1}{\pi^2/16} = 16\cdot\frac{\pi/2 - 1}{\pi^2} = \frac{8\pi - 16}{\pi^2}$ , about $0.9254$ . The order matters: $\frac{\tan x - x\sec^2 x}{x^2}$ gives $-0.9254$ , a slope of the wrong sign.

d) $u = \sec x$ , $u' = \sec x\tan x$ ; $v = \tan x$ , $v' = \sec^2 x$ . Numerator: $\sec x\tan x\cdot\tan x - \sec x\cdot\sec^2 x = \sec x\left(\tan^2 x - \sec^2 x\right) = -\sec x$ , by $\sec^2 x = 1 + \tan^2 x$ . So $h'(x) = \frac{-\sec x}{\tan^2 x} = -\frac{1}{\cos x}\cdot\frac{\cos^2 x}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\csc x\cot x$ . Second road: $h(x) = \frac{1}{\cos x}\cdot\frac{\cos x}{\sin x} = \frac{1}{\sin x} = \csc x$ , so $h'(x) = -\csc x\cot x$ at once. But $h$ is NOT defined where $\cos x = 0$ , although $\csc x$ is: the formula holds only where $\sin x \ne 0$ and $\cos x \ne 0$ . At $\frac{\pi}{3}$ : $-\frac{2}{\sqrt 3}\cdot\frac{1}{\sqrt 3} = -\frac{2}{3}$ .

e) Quotient rule, $u = x^2$ , $v = \cos x$ : $s'(x) = \frac{2x\cos x - x^2(-\sin x)}{\cos^2 x} = \frac{2x\cos x + x^2\sin x}{\cos^2 x}$ . Product rule on $x^2\sec x$ : $s'(x) = 2x\sec x + x^2\sec x\tan x$ . They agree: split the fraction, $\frac{2x\cos x}{\cos^2 x} = \frac{2x}{\cos x} = 2x\sec x$ and $\frac{x^2\sin x}{\cos^2 x} = x^2\cdot\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = x^2\sec x\tan x$ . At $\frac{\pi}{3}$ , $\sec = 2$ , $\tan = \sqrt 3$ : $s' = 2\cdot\frac{\pi}{3}\cdot 2 + \frac{\pi^2}{9}\cdot 2\sqrt 3 = \frac{4\pi}{3} + \frac{2\sqrt 3\,\pi^2}{9}$ , about $7.9876$ . When the denominator is a single $\cos x$ , the product with $\sec x$ is usually shorter.

Exercise 4: Tangent and normal lines with exact trigonometric values

The tangent to $y = f(x)$ at $a$ is $y = f(a) + f'(a)(x - a)$ , and the normal is the line through the same point with slope $-\frac{1}{f'(a)}$ when $f'(a) \ne 0$ . Every value is exact: $\csc\frac{\pi}{6} = 2$ , $\sec\frac{\pi}{4} = \sqrt 2$ , and $\pi$ stays $\pi$ in the equation. The calculator is only used to CHECK a decimal, in radian mode.

The figure shows $y = x\sin x$ for $0 \le x \le 8.6$ , between the dashed lines $y = x$ and $y = -x$ , and the three points where it touches them; $P$ is the first, at $x = \frac{\pi}{2}$ .

a) Find the tangent to $y = \csc x$ at $x = \frac{\pi}{6}$ , in the form $y = mx + b$ .
b) Find the tangent and the normal to $y = x\sin x$ at $P$ . Show that the normal meets the curve again on the $x$ -axis.
c) Show that at EVERY point where $\sin x = 1$ the tangent to $y = x\sin x$ is the line $y = x$ , and at every point where $\sin x = -1$ it is $y = -x$ . What does the figure show there?
d) Find the tangent and the normal to $y = \sec x$ at $x = \frac{\pi}{4}$ .
e) Find the tangent to $y = \frac{\cos x}{x}$ at $x = \frac{\pi}{2}$ and the point where it crosses the $y$ -axis.

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Answers

a) $y = -2\sqrt 3\,x + 2 + \frac{\sqrt 3\,\pi}{3}$
b) Tangent $y = x$ ; normal $y = \pi - x$ , which meets the curve at $(\pi, 0)$ .
c) At $\sin a = 1$ : $y(a) = a$ , $y'(a) = 1$ , tangent $y = x$ ; at $\sin a = -1$ : tangent $y = -x$ . The curve touches the dashed lines there.
d) Tangent $y = \sqrt 2 + \sqrt 2\left(x - \frac{\pi}{4}\right)$ ; normal $y = \sqrt 2 - \frac{\sqrt 2}{2}\left(x - \frac{\pi}{4}\right)$
e) $y = 1 - \frac{2x}{\pi}$ , crossing the $y$ -axis at $(0, 1)$

a) Point: $\csc\frac{\pi}{6} = \frac{1}{\sin(\pi/6)} = 2$ . Slope: $(\csc x)' = -\csc x\cot x$ , and $\cot\frac{\pi}{6} = \sqrt 3$ , so $m = -2\sqrt 3$ . Tangent: $y = 2 - 2\sqrt 3\left(x - \frac{\pi}{6}\right) = -2\sqrt 3\,x + 2 + \frac{2\sqrt 3\,\pi}{6} = -2\sqrt 3\,x + 2 + \frac{\sqrt 3\,\pi}{3}$ . The intercept is about $3.8138$ . Expanding $-2\sqrt 3\cdot\left(-\frac{\pi}{6}\right)$ is where the sign goes: minus times minus is plus.

b) $y(\frac{\pi}{2}) = \frac{\pi}{2}\sin\frac{\pi}{2} = \frac{\pi}{2}$ . Product rule: $y' = \sin x + x\cos x$ , so $y'\left(\frac{\pi}{2}\right) = 1 + \frac{\pi}{2}\cdot 0 = 1$ . Tangent: $y = \frac{\pi}{2} + \left(x - \frac{\pi}{2}\right) = x$ . Normal: slope $-\frac{1}{1} = -1$ , so $y = \frac{\pi}{2} - \left(x - \frac{\pi}{2}\right) = \pi - x$ . It meets the $x$ -axis at $(\pi, 0)$ , and $\pi\sin\pi = 0$ : that point is on the curve. The solution figure is drawn with equal scales, so the right angle between tangent and normal is true.

c) Let $a$ be a point where $\sin a = 1$ ; then $\cos a = 0$ , by $\sin^2 a + \cos^2 a = 1$ . So $y(a) = a\cdot 1 = a$ and $y'(a) = \sin a + a\cos a = 1 + 0 = 1$ : the tangent is $y = a + 1\cdot(x - a) = x$ , whatever $a$ is. Where $\sin a = -1$ , again $\cos a = 0$ , $y(a) = -a$ , $y'(a) = -1$ , and the tangent is $y = -a - (x - a) = -x$ . Since $|x\sin x| \le |x|$ , the curve lies between the two dashed lines, and at those points it touches them, tangentially, without crossing: the orange dots of the figure. The key step is $\cos a = 0$ , which kills the term $a\cos a$ that would otherwise depend on $a$ .

d) $\sec\frac{\pi}{4} = \frac{1}{\cos(\pi/4)} = \frac{2}{\sqrt 2} = \sqrt 2$ and $\tan\frac{\pi}{4} = 1$ . Slope: $\sec x\tan x = \sqrt 2$ . Tangent: $y = \sqrt 2 + \sqrt 2\left(x - \frac{\pi}{4}\right)$ . Normal slope: $-\frac{1}{\sqrt 2} = -\frac{\sqrt 2}{2}$ , after rationalizing; normal: $y = \sqrt 2 - \frac{\sqrt 2}{2}\left(x - \frac{\pi}{4}\right)$ . The normal slope is the NEGATIVE reciprocal: $\frac{1}{\sqrt 2}$ alone gives a line that is not perpendicular to anything useful.

e) $y\left(\frac{\pi}{2}\right) = \frac{\cos(\pi/2)}{\pi/2} = 0$ . Quotient rule, $u = \cos x$ , $u' = -\sin x$ , $v = x$ , $v' = 1$ : $y' = \frac{-x\sin x - \cos x}{x^2}$ , so $y'\left(\frac{\pi}{2}\right) = \frac{-\frac{\pi}{2}\cdot 1 - 0}{\pi^2/4} = -\frac{\pi}{2}\cdot\frac{4}{\pi^2} = -\frac{2}{\pi}$ , about $-0.6366$ . Tangent: $y = 0 - \frac{2}{\pi}\left(x - \frac{\pi}{2}\right) = 1 - \frac{2x}{\pi}$ . At $x = 0$ it gives $y = 1$ : it crosses the $y$ -axis at $(0, 1)$ . Dividing by the fraction $\frac{\pi^2}{4}$ is multiplying by $\frac{4}{\pi^2}$ : the complex fraction is where this slope is usually lost.

Exercise 5: Higher derivatives by the cycle of four, and the equation y'' + y = 0

The derivatives of $\sin x$ repeat every four steps, $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$ , and so do those of $\cos x$ . A derivative of order $n$ is read off the REMAINDER of $n$ divided by $4$ .

The same two derivatives make $\sin x$ and $\cos x$ solutions of $y'' + y = 0$ , the equation of every oscillation of this chapter. Nothing is solved here: each claim is VERIFIED by differentiating twice and substituting.

a) Find $\frac{d^{35}}{dx^{35}}\cos x$ and $\frac{d^{102}}{dx^{102}}\sin x$ .
b) Let $f(x) = 3\sin x - 5\cos x$ . Find $f(0)$ , $f'(0)$ , $f''(0)$ , $f'''(0)$ , then $f^{(2026)}(0)$ and $f^{(2027)}(0)$ .
c) Show that $y = A\cos x + B\sin x$ satisfies $y'' + y = 0$ for all constants $A$ and $B$ . Find the constants for which $y(0) = 2$ and $y'(0) = -3$ .
d) Show that $y = x\sin x$ satisfies $y'' + y = 2\cos x$ . Deduce the constant $A$ for which $y = Ax\sin x$ satisfies $y'' + y = 6\cos x$ .
e) Compute $y''$ for $y = \csc x$ and decide whether $y = \csc x$ satisfies $y'' + y = 0$ . Evaluate $y''\left(\frac{\pi}{2}\right)$ .

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Answers

a) $\sin x$ and $-\sin x$
b) $-5, 3, 5, -3$ repeating; $f^{(2026)}(0) = 5$ , $f^{(2027)}(0) = -3$
c) $y'' = -y$ ; $A = 2$ , $B = -3$ : $y = 2\cos x - 3\sin x$
d) $y'' = 2\cos x - x\sin x$ , so $y'' + y = 2\cos x$ ; $A = 3$
e) $y'' = \csc x\left(\cot^2 x + \csc^2 x\right)$ , $y'' + y = 2\csc^3 x \ne 0$ : no; $y''\left(\frac{\pi}{2}\right) = 1$

a) The derivatives of $\cos x$ are $-\sin x$ , $-\cos x$ , $\sin x$ , $\cos x$ , then again. $35 = 4\times 8 + 3$ , so $\frac{d^{35}}{dx^{35}}\cos x = (\cos x)''' = \sin x$ . For sine, $102 = 4\times 25 + 2$ , so $\frac{d^{102}}{dx^{102}}\sin x = (\sin x)'' = -\sin x$ . Check each division by multiplying back: $32 + 3 = 35$ and $100 + 2 = 102$ . The parity of $n$ only says sine or cosine; the sign needs the remainder.

b) $f(0) = 0 - 5 = -5$ . $f'(x) = 3\cos x + 5\sin x$ , $f'(0) = 3$ . $f''(x) = -3\sin x + 5\cos x$ , $f''(0) = 5$ . $f'''(x) = -3\cos x - 5\sin x$ , $f'''(0) = -3$ . $f^{(4)}(x) = 3\sin x - 5\cos x = f(x)$ , so the values at $0$ repeat: $-5, 3, 5, -3, -5, \dots$ Now $2026 = 4\times 506 + 2$ , so $f^{(2026)}(0) = f''(0) = 5$ , and $2027 = 4\times 506 + 3$ , so $f^{(2027)}(0) = f'''(0) = -3$ . The minus in front of $5\cos x$ is carried through every step: $(-5\cos x)' = +5\sin x$ .

c) $y' = -A\sin x + B\cos x$ and $y'' = -A\cos x - B\sin x = -y$ , so $y'' + y = 0$ for every $A$ and $B$ : a VERIFICATION by two differentiations. Then $y(0) = A\cos 0 + B\sin 0 = A$ , so $A = 2$ , and $y'(0) = -A\cdot 0 + B\cdot 1 = B$ , so $B = -3$ . The function is $y = 2\cos x - 3\sin x$ . The value at $0$ fixes the cosine coefficient, the slope at $0$ fixes the sine coefficient.

d) Product rule: $y' = \sin x + x\cos x$ . Again, on each term: $y'' = \cos x + (\cos x - x\sin x) = 2\cos x - x\sin x$ . So $y'' + y = 2\cos x - x\sin x + x\sin x = 2\cos x$ . For $y = Ax\sin x$ , the constant multiple rule multiplies everything by $A$ : $y'' + y = 2A\cos x$ , and this equals $6\cos x$ for every $x$ exactly when $A = 3$ . The term $x\sin x$ cancels, which is the reason this function is chosen: it produces a pure cosine.

e) $y' = -\csc x\cot x$ . By Exercise 2 e), $(\csc x\cot x)' = -\csc x(\cot^2 x + \csc^2 x)$ , so $y'' = \csc x\left(\cot^2 x + \csc^2 x\right)$ . Then $y'' + y = \csc x\left(\cot^2 x + 1 + \csc^2 x\right) = \csc x\cdot 2\csc^2 x = 2\csc^3 x$ , using $\cot^2 x + 1 = \csc^2 x$ . This is never $0$ , because $\csc x$ never is: $\csc x$ does NOT satisfy $y'' + y = 0$ . At $\frac{\pi}{2}$ : $\csc = 1$ , $\cot = 0$ , $y'' = 1$ and $y'' + y = 2$ . Being built from $\sin x$ is not enough; only the combinations $A\cos x + B\sin x$ pass.

Exercise 6: Horizontal and parallel tangents on an interval: one function, factor, every quadrant

A horizontal tangent at $a$ means $f'(a) = 0$ ; a tangent parallel to a line of slope $m$ means $f'(a) = m$ . With trigonometric functions both are trigonometric EQUATIONS, and three algebra gestures decide them: turn the equation into ONE function with an identity, FACTOR instead of dividing, and read EVERY solution of the interval on the unit circle, not only the one that $\cos^{-1}$ or $\sin^{-1}$ returns.

The figure shows $g(x) = \sin x\,(1 - \cos x)$ on $[0, 2\pi]$ , the function of question b).

a) Find the point of $f(x) = 2\sec x - \tan x$ , $-\frac{\pi}{2} < x < \frac{\pi}{2}$ , where the tangent is horizontal.
b) Find every point of $g(x) = \sin x\,(1 - \cos x)$ , $0 \le x \le 2\pi$ , where the tangent is horizontal, using $\sin^2 x = 1 - \cos^2 x$ to get an equation in $\cos x$ alone. How many are there?
c) Find the points of $y = \tan x$ , $-\frac{\pi}{2} < x < \frac{\pi}{2}$ , where the tangent is parallel to the line $y = 4x$ .
d) Find every $x$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ where the tangents to $y = \sin x$ and $y = \tan x$ are parallel.
e) Find every point of $f(x) = \sin x - x\cos x$ , $0 \le x \le 2\pi$ , where the tangent is horizontal.

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Answers

a) $\left(\frac{\pi}{6}, \sqrt 3\right)$
b) Four points: $(0, 0)$ , $\left(\frac{2\pi}{3}, \frac{3\sqrt 3}{4}\right)$ , $\left(\frac{4\pi}{3}, -\frac{3\sqrt 3}{4}\right)$ , $(2\pi, 0)$
c) $\left(\frac{\pi}{3}, \sqrt 3\right)$ and $\left(-\frac{\pi}{3}, -\sqrt 3\right)$
d) Only $x = 0$ ( $\cos^3 x = 1$ ), where both tangents are $y = x$ .
e) $(0, 0)$ , $(\pi, \pi)$ , $(2\pi, -2\pi)$

a) $f'(x) = 2\sec x\tan x - \sec^2 x = \sec x\,(2\tan x - \sec x)$ , factoring the common $\sec x$ . The factor $\sec x = \frac{1}{\cos x}$ is never $0$ , so $f'(x) = 0 \iff 2\tan x = \sec x$ . Multiply by $\cos x$ , which is positive on the interval, so no solution is created or lost: $2\sin x = 1$ , $\sin x = \frac{1}{2}$ . On $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ the only solution is $x = \frac{\pi}{6}$ ; the other angle with sine $\frac{1}{2}$ , $\frac{5\pi}{6}$ , is outside the interval. Height: $f\left(\frac{\pi}{6}\right) = 2\cdot\frac{2}{\sqrt 3} - \frac{1}{\sqrt 3} = \frac{3}{\sqrt 3} = \sqrt 3$ . The complex equation $2\tan x = \sec x$ became a basic one by clearing the denominators $\cos x$ .

b) Product rule: $g'(x) = \cos x\,(1 - \cos x) + \sin x\,(\sin x) = \cos x - \cos^2 x + \sin^2 x$ . Replace $\sin^2 x$ by $1 - \cos^2 x$ to get ONE function: $g'(x) = -2\cos^2 x + \cos x + 1$ , a quadratic in $c = \cos x$ , which factors as $-(2c + 1)(c - 1)$ . So $g'(x) = 0 \iff \cos x = -\frac{1}{2}$ or $\cos x = 1$ : $x = \frac{2\pi}{3}$ , $x = \frac{4\pi}{3}$ , and $x = 0$ , $x = 2\pi$ , the endpoints of the closed interval. Heights: $g\left(\frac{2\pi}{3}\right) = \frac{\sqrt 3}{2}\cdot\frac{3}{2} = \frac{3\sqrt 3}{4}$ , $g\left(\frac{4\pi}{3}\right) = -\frac{3\sqrt 3}{4}$ , $g(0) = g(2\pi) = 0$ . Four points. The two at the endpoints are the ones that get lost: the factor $\cos x - 1$ vanishes only there, and the curve leaves the axis FLAT without turning back, as the figure shows. A horizontal tangent is not necessarily a peak or a valley. The derivative of $-\cos x$ is $+\sin x$ , which is where the plus in $\sin x\,(\sin x)$ comes from.

c) Parallel to $y = 4x$ means slope $4$ : $\sec^2 x = 4$ , so $\cos^2 x = \frac{1}{4}$ and $\cos x = \pm\frac{1}{2}$ . On $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ cosine is positive, so only $\cos x = \frac{1}{2}$ is possible, and it has TWO solutions there: $x = \frac{\pi}{3}$ and $x = -\frac{\pi}{3}$ . Points: $\left(\frac{\pi}{3}, \sqrt 3\right)$ and $\left(-\frac{\pi}{3}, -\sqrt 3\right)$ . Two losses are possible here: the $\pm$ of the square root, which is rejected for a REASON (the sign of cosine on the interval), and the negative angle, which $\cos^{-1}$ never returns.

d) Equal slopes: $\cos x = \sec^2 x = \frac{1}{\cos^2 x}$ . Multiplying by $\cos^2 x \ne 0$ : $\cos^3 x = 1$ , so $\cos x = 1$ , because the only real cube root of $1$ is $1$ . On the interval, $x = 0$ . At $0$ both curves pass through the origin with slope $1$ : their tangents are the SAME line, $y = x$ . Everywhere else in the interval $0 < \cos x < 1 < \sec^2 x$ , so the tangent to $\tan x$ is always steeper.

e) Product rule on $x\cos x$ , then subtract the WHOLE derivative: $f'(x) = \cos x - \left(\cos x - x\sin x\right) = x\sin x$ . Forgetting the bracket gives $\cos x - \cos x - x\sin x = -x\sin x$ , which happens to have the same zeros but the wrong sign everywhere. Now $x\sin x = 0 \iff x = 0$ or $\sin x = 0$ , that is $x = 0$ , $\pi$ , $2\pi$ on $[0, 2\pi]$ . Heights: $f(0) = 0$ , $f(\pi) = 0 - \pi(-1) = \pi$ , $f(2\pi) = 0 - 2\pi\cdot 1 = -2\pi$ . Dividing by $x$ would lose the endpoint $x = 0$ , where the tangent is indeed horizontal.

Exercise 7: Reading f and f' on one figure: secant, tangent and their derivatives

The figure shows three curves $A$ , $B$ and $C$ on $-1.25 \le x \le 1.25$ , inside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . They are the graphs of $y = \sec x$ , $y = \tan x$ and $y = \sec x\tan x$ , in some order.

Two facts decide most of the questions without any computation: the derivative of an EVEN function is odd, the derivative of an ODD function is even, and where a curve has a horizontal tangent its derivative crosses zero.

a) Identify the curve of $y = \sec x$ , of $y = \tan x$ and of $y = \sec x\tan x$ , with a reason for each.
b) One of the three curves is the derivative of another: which, and how does the figure confirm it? Of which curve is the derivative equal to the SQUARE of another?
c) Compute the slopes of $B$ and of $C$ at $x = 0$ , the second by the product rule. What do you notice about their tangent lines at the origin?
d) Find, by factoring, the point where $A$ and $C$ cross.
e) Using parity only, give the slope of $A$ at $x = -\frac{\pi}{4}$ from its slope at $\frac{\pi}{4}$ , then confirm by a direct computation.

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Answers

a) $A$ is $\sec x$ (even, value $1$ at $0$ ), $B$ is $\tan x$ , $C$ is $\sec x\tan x$ (odd, above $B$ for $x > 0$ ).
b) $C = A'$ , since $(\sec x)' = \sec x\tan x$ ; and $B' = \sec^2 x = A^2$ .
c) Both slopes are $1$ : $B$ and $C$ share the tangent $y = x$ at the origin.
d) $\sec x\,(\tan x - 1) = 0$ : $\left(\frac{\pi}{4}, \sqrt 2\right)$
e) $-\sqrt 2$

a) At $x = 0$ : $\sec 0 = 1$ , $\tan 0 = 0$ , $\sec 0\tan 0 = 0$ . The only curve through $(0, 1)$ is $A$ , which is also symmetric about the $y$ -axis, as $\sec x = \frac{1}{\cos x}$ is even: $A$ is $\sec x$ . $B$ and $C$ both pass through the origin and are odd. To separate them, compare at $\frac{\pi}{4}$ : $\tan\frac{\pi}{4} = 1$ while $\sec\frac{\pi}{4}\tan\frac{\pi}{4} = \sqrt 2 > 1$ . For $0 < x < \frac{\pi}{2}$ , $\sec x > 1$ , so $\sec x\tan x > \tan x$ : the steeper curve $C$ is $\sec x\tan x$ , and $B$ is $\tan x$ .

b) $(\sec x)' = \sec x\tan x$ : $C$ is the derivative of $A$ . The figure agrees: $A$ is flat at its lowest point $x = 0$ , exactly where $C$ crosses zero; to the left $A$ falls and $C$ is below the axis, to the right $A$ rises and $C$ is above; the steeper $A$ gets near the edges, the larger $|C|$ is. And $(\tan x)' = \sec^2 x = (\sec x)^2$ : the derivative of $B$ is the square of $A$ . An even function ( $A$ ) has an odd derivative ( $C$ ), an odd function ( $B$ ) an even one ( $A^2$ ).

c) $B$ : $(\tan x)' = \sec^2 x$ , which is $1$ at $0$ . $C$ : product rule with $u = \sec x$ , $v = \tan x$ : $(\sec x\tan x)' = \sec x\tan x\cdot\tan x + \sec x\cdot\sec^2 x = \sec x\tan^2 x + \sec^3 x$ , which is $0 + 1 = 1$ at $0$ . Both curves pass through the origin with slope $1$ : they have the SAME tangent line there, $y = x$ , and only separate further out, $C$ bending up faster. On the figure the two curves are practically indistinguishable near the origin.

d) $\sec x = \sec x\tan x \iff \sec x\tan x - \sec x = 0 \iff \sec x\,(\tan x - 1) = 0$ . The factor $\sec x$ is never $0$ , so $\tan x = 1$ , and on the interval $x = \frac{\pi}{4}$ , where $y = \sec\frac{\pi}{4} = \sqrt 2$ . The crossing point is $\left(\frac{\pi}{4}, \sqrt 2\right)$ , about $(0.785, 1.414)$ on the figure. Here dividing by $\sec x$ would be legitimate, but only because it never vanishes, and the factored form is what shows it.

e) $A$ is even, so its derivative is odd: the slope at $-\frac{\pi}{4}$ is the opposite of the slope at $\frac{\pi}{4}$ . At $\frac{\pi}{4}$ the slope is $\sec\frac{\pi}{4}\tan\frac{\pi}{4} = \sqrt 2$ , so at $-\frac{\pi}{4}$ it is $-\sqrt 2$ . Directly: $\sec\left(-\frac{\pi}{4}\right) = \sqrt 2$ and $\tan\left(-\frac{\pi}{4}\right) = -1$ , product $-\sqrt 2$ . Symmetry is a free check on any derivative of an even or odd function.

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, support the correction with a counterexample or a computation, and write the correct statement. A scientific calculator is allowed, as on the exam.

a) $(\cot x)' = -\csc x\cot x$ , the same pattern as $(\csc x)'$ .
b) $(\tan x)'' = (\sec^2 x)' = 2\sec x$ , by the power rule.
c) With my calculator, the slope of $y = \sin x$ at $x = 2$ is $\cos 2 = 0.9994$ .
d) For $y = \tan x - 2x$ on $[0, 2\pi]$ : $y' = \sec^2 x - 2 = 0$ gives $\cos^2 x = \frac{1}{2}$ , so $\cos x = \frac{\sqrt 2}{2}$ , and my calculator gives $x = 0.7854$ . There is one horizontal tangent.
e) $\frac{\sin x + \tan x}{\sin x} = 1 + \tan x$ , so its derivative is $\sec^2 x$ .

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

a)

The statement is

True slope of

\cot x

at

\frac{\pi}{2}

b)

The statement is

(\tan x)''

at

x = \frac{\pi}{4}

c)

The statement is

The true slope at

x = 2

d)

The statement is

Number of horizontal tangents

e)

The statement is

True slope at

\frac{\pi}{4}

Show the solution

Answers

a) False: $(\cot x)' = -\csc^2 x$ .
b) False: $\sec^2 x = \sec x\cdot\sec x$ , and $(\tan x)'' = 2\sec^2 x\tan x$ .
c) False: the calculator was in degrees; $\cos 2 \approx -0.4161$ in radians.
d) False: $\cos x = \pm\frac{\sqrt 2}{2}$ gives four points, $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ .
e) False: $\frac{\tan x}{\sin x} = \sec x$ , so the function is $1 + \sec x$ and its derivative $\sec x\tan x$ .

a) FALSE. $\cot x = \frac{\cos x}{\sin x}$ , and the quotient rule gives $(\cot x)' = \frac{-\sin x\cdot\sin x - \cos x\cdot\cos x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x$ . Counterexample to the student's formula: at $\frac{\pi}{2}$ it gives $-\csc\frac{\pi}{2}\cot\frac{\pi}{2} = -1\cdot 0 = 0$ , a horizontal tangent, while $\cot x$ crosses the axis there with slope $-\csc^2\frac{\pi}{2} = -1$ . Correct statement: $(\cot x)' = -\csc^2 x$ pairs with $(\tan x)' = \sec^2 x$ , and $(\csc x)' = -\csc x\cot x$ pairs with $(\sec x)' = \sec x\tan x$ . The pairs never cross.

b) FALSE. The power rule is about $x^n$ , a power of the VARIABLE; $\sec^2 x$ is a power of a function. Before the chain rule, write it as a product: $(\sec x\cdot\sec x)' = \sec x\tan x\cdot\sec x + \sec x\cdot\sec x\tan x = 2\sec^2 x\tan x$ . Counterexample: at $x = 0$ the student gets $2\sec 0 = 2$ , but $\tan x$ is odd, so $(\tan x)'$ is even and $(\tan x)''$ is odd, hence $0$ at $0$ ; indeed $2\sec^2 0\tan 0 = 0$ . Correct statement: $(\tan x)'' = 2\sec^2 x\tan x$ .

c) FALSE. $0.9994$ is $\cos(2^\circ)$ : the calculator was in DEGREE mode. Every formula of the chapter, $(\sin x)' = \cos x$ included, assumes $x$ in radians. In radian mode, $\cos 2 \approx -0.4161$ . Sanity check without any calculator: $2$ is between $\frac{\pi}{2} \approx 1.57$ and $\pi$ , where sine is coming down from its top, so the slope must be NEGATIVE. Correct statement: the slope of $y = \sin x$ at $x = 2$ is $\cos 2 \approx -0.4161$ , calculator in radians. Set the mode before the exam starts, and test it: $\cos\pi$ must display $-1$ .

d) FALSE. From $\cos^2 x = \frac{1}{2}$ the square root gives $\cos x = \pm\frac{\sqrt 2}{2}$ , not only the positive value, and on $[0, 2\pi]$ each value is taken TWICE. $\cos x = \frac{\sqrt 2}{2}$ at $\frac{\pi}{4}$ and $\frac{7\pi}{4}$ ; $\cos x = -\frac{\sqrt 2}{2}$ at $\frac{3\pi}{4}$ and $\frac{5\pi}{4}$ . None of the four is an asymptote $\frac{\pi}{2}$ or $\frac{3\pi}{2}$ , so all are in the domain. The calculator's $\cos^{-1}$ returns ONE angle, in $[0, \pi]$ : the others come from the unit circle. Correct statement: $y = \tan x - 2x$ has four horizontal tangents on $[0, 2\pi]$ , at $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ .

e) FALSE. Dividing a sum by $\sin x$ divides EACH term: $\frac{\sin x}{\sin x} + \frac{\tan x}{\sin x} = 1 + \frac{\sin x}{\cos x}\cdot\frac{1}{\sin x} = 1 + \sec x$ , for $\sin x \ne 0$ , $\cos x \ne 0$ . The student cancelled $\sin x$ against one term only. The derivative is $\sec x\tan x$ . Counterexample at $\frac{\pi}{4}$ : the value of the function is $\frac{\sqrt 2/2 + 1}{\sqrt 2/2} = 1 + \sqrt 2$ , not $1 + \tan\frac{\pi}{4} = 2$ ; the true slope is $\sqrt 2$ , not $2$ . Correct statement: $\frac{\sin x + \tan x}{\sin x} = 1 + \sec x$ , with derivative $\sec x\tan x$ .

Exercise 9: A Ferris wheel: height, vertical velocity and the equation y'' + y = 0

A Ferris wheel of radius $20$ m has its centre $C$ at $22$ m above the ground. It turns at a constant $1$ radian per minute, so one turn takes $2\pi \approx 6.28$ minutes. A rider boards at the lowest point $B$ at time $t = 0$ . After $t$ minutes the angle $BCP$ between the downward vertical and the rider's position $P$ is $t$ radians, as in the figure.

The rider's height above the ground and horizontal position, measured from the vertical line through $C$ , are then $h(t) = 22 - 20\cos t$ and $x(t) = 20\sin t$ , in metres. Give exact values where they are natural; when a decimal is asked, round as stated, calculator in RADIAN mode.

a) Find $h(0)$ , the vertical velocity $h'(t)$ and $h'\left(\frac{\pi}{6}\right)$ , with units. What does the sign of $h'\left(\frac{\pi}{6}\right)$ say?
b) Find $h''(t)$ . Show that $y(t) = h(t) - 22$ satisfies $y'' + y = 0$ , and say what $y$ measures.
c) During the first turn, when is the rider rising fastest, and at what height? When is the vertical velocity zero, and where is the rider then?
d) Find the horizontal velocity $x'(t)$ and show that the speed $\sqrt{x'(t)^2 + h'(t)^2}$ is constant. Explain its value.
e) At $t = 2$ min, find $h(2)$ , $h'(2)$ and $h''(2)$ to two decimals. Is the rider going up or down, and is the vertical velocity increasing or decreasing? What would a calculator in degree mode give for $h(2)$ ?

Show the solution

Answers

a) $h(0) = 2$ m; $h'(t) = 20\sin t$ m/min; $h'\left(\frac{\pi}{6}\right) = 10$ m/min, rising.
b) $h''(t) = 20\cos t = -(h - 22)$ , so $y'' = -y$ ; $y$ is the height above the centre.
c) Fastest rise $20$ m/min at $t = \frac{\pi}{2} \approx 1.57$ min, at $22$ m; $h' = 0$ at $t = 0$ (bottom, $2$ m) and $t = \pi$ (top, $42$ m).
d) $x'(t) = 20\cos t$ ; speed $= 20\sqrt{\cos^2 t + \sin^2 t} = 20$ m/min $= 20$ m $\times\, 1$ rad/min.
e) $h(2) \approx 30.32$ m, $h'(2) \approx 18.19$ m/min, $h''(2) \approx -8.32$ m/min²: going up, vertical velocity decreasing. Degree mode: $2.01$ m.

a) $h(0) = 22 - 20\cos 0 = 2$ m: the lowest seat is $2$ m above the ground. By the constant multiple rule, $h'(t) = -20(-\sin t) = 20\sin t$ , in metres per minute. At $\frac{\pi}{6}$ : $h' = 20\cdot\frac{1}{2} = 10$ m/min. It is positive: the rider is rising at $10$ m per minute half a radian after boarding. Once more the minus in front of $\cos t$ meets the minus of $(\cos t)'$ .

b) $h''(t) = 20\cos t$ , in metres per minute per minute. With $y = h - 22 = -20\cos t$ : $y'' = h'' = 20\cos t = -y$ , so $y'' + y = 0$ . The quantity $y$ is the height of the rider ABOVE THE CENTRE of the wheel, negative on the lower half. The constant $22$ disappears in $y''$ but not in $y$ : that is why $h$ itself does not satisfy $h'' + h = 0$ , since $h'' + h = 22$ . Only the displacement from the centre oscillates in the pure form of this chapter.

c) $h'(t) = 20\sin t \le 20$ , with equality exactly when $\sin t = 1$ , that is $t = \frac{\pi}{2} \approx 1.57$ min in the first turn. The rider then rises at $20$ m/min, at height $h\left(\frac{\pi}{2}\right) = 22 - 0 = 22$ m, level with the centre, on the rising side of the wheel. No extremum theorem is needed: the bound $|\sin t| \le 1$ settles it. The vertical velocity is zero when $\sin t = 0$ : $t = 0$ , at the bottom, $2$ m, and $t = \pi \approx 3.14$ min, at the top, $h(\pi) = 22 + 20 = 42$ m. At both ends of the diameter the rider moves horizontally.

d) $x'(t) = 20\cos t$ m/min. Then $x'(t)^2 + h'(t)^2 = 400\cos^2 t + 400\sin^2 t = 400(\cos^2 t + \sin^2 t) = 400$ , and the speed is $\sqrt{400} = 20$ m/min at every instant. It is the radius times the angular rate: $20$ m $\times 1$ rad/min, the length of arc covered per minute. The Pythagorean identity is exactly the statement that a point moving on a circle at a constant angular rate has a constant speed; only the split between its horizontal and vertical parts changes.

e) Radian mode: $\cos 2 \approx -0.41615$ and $\sin 2 \approx 0.90930$ . So $h(2) = 22 - 20\cos 2 \approx 30.32$ m, $h'(2) = 20\sin 2 \approx 18.19$ m/min, $h''(2) = 20\cos 2 \approx -8.32$ m/min². Since $2 > \frac{\pi}{2}$ , the rider is on the upper half, past the level of the centre. $h'(2) > 0$ : still going UP. $h''(2) < 0$ : the vertical velocity is DECREASING, it will reach $0$ at the top, at $t = \pi$ . In degree mode the calculator computes $\cos(2^\circ) \approx 0.99939$ and gives $h \approx 2.01$ m, a rider who has barely left the boarding platform after two minutes: an answer that the picture refutes at once.

Exercise 10: A final exam question: the function sec x + tan x

Let $f(x) = \sec x + \tan x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ; its graph is shown in the figure. This is the shape of a long final exam question: every step is an identity or a rule, named, and every value is exact. The whole question turns on one algebraic gesture, the conjugate $1 - \sin x$ .

a) Show that $f(x) = \frac{1 + \sin x}{\cos x}$ , then, multiplying by the conjugate, that $f(x) = \frac{\cos x}{1 - \sin x}$ . Deduce that $f(x) > 0$ on the interval, and give $f(0)$ .
b) Compute $f'(x)$ and show that $f'(x) = \sec x\cdot f(x)$ . Find $f'(0)$ .
c) Deduce from a) and b) that $f'(x) = \frac{1}{1 - \sin x}$ . Show that $f'(x) > \frac{1}{2}$ on the interval: can the graph have a horizontal tangent?
d) Find $f''(x)$ from the form of c), and show that $f''(x) = f(x)\,f'(x)$ . Find $f''(0)$ .
e) Find the point where the tangent has slope $2$ , the equation of that tangent, and its $y$ -intercept.

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Answers

a) $f(x) = \frac{1 + \sin x}{\cos x} = \frac{\cos x}{1 - \sin x} > 0$ ; $f(0) = 1$
b) $f'(x) = \sec x\tan x + \sec^2 x = \sec x\,(\tan x + \sec x)$ ; $f'(0) = 1$
c) $f'(x) = \frac{1}{1 - \sin x} > \frac{1}{2}$ : no horizontal tangent.
d) $f''(x) = \frac{\cos x}{(1 - \sin x)^2} = f(x)f'(x)$ ; $f''(0) = 1$
e) $\left(\frac{\pi}{6}, \sqrt 3\right)$ ; $y = 2x + \sqrt 3 - \frac{\pi}{3}$ , intercept $\sqrt 3 - \frac{\pi}{3}$

a) $\sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x}$ . Multiply top and bottom by the conjugate $1 - \sin x$ , which is positive on the interval because $\sin x < 1$ there: $\frac{(1 + \sin x)(1 - \sin x)}{\cos x\,(1 - \sin x)} = \frac{1 - \sin^2 x}{\cos x\,(1 - \sin x)} = \frac{\cos^2 x}{\cos x\,(1 - \sin x)} = \frac{\cos x}{1 - \sin x}$ , cancelling the factor $\cos x \ne 0$ . On $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , $\cos x > 0$ and $1 + \sin x > 0$ , so $f(x) > 0$ . And $f(0) = \frac{1 + 0}{1} = 1$ . The conjugate turns $1 + \sin x$ into $\cos^2 x$ : the same gesture as with a square root.

b) By the sum rule, $f'(x) = \sec x\tan x + \sec^2 x$ . Factor the common $\sec x$ : $f'(x) = \sec x\,(\tan x + \sec x) = \sec x\cdot f(x)$ . At $0$ : $f'(0) = 1\cdot 1 = 1$ . Factoring is what reveals the function inside its own derivative.

c) With the second form of a): $f'(x) = \frac{1}{\cos x}\cdot\frac{\cos x}{1 - \sin x} = \frac{1}{1 - \sin x}$ , cancelling $\cos x \ne 0$ . On the interval $-1 < \sin x < 1$ , so $0 < 1 - \sin x < 2$ and $f'(x) = \frac{1}{1 - \sin x} > \frac{1}{2}$ . The derivative is never $0$ : the graph has no horizontal tangent, every tangent line rises with slope greater than $\frac{1}{2}$ . Check at $0$ : $\frac{1}{1 - 0} = 1$ , as in b).

d) Quotient rule on $\frac{1}{1 - \sin x}$ , with $u = 1$ , $u' = 0$ , $v = 1 - \sin x$ , $v' = -\cos x$ : $f''(x) = \frac{0\cdot(1 - \sin x) - 1\cdot(-\cos x)}{(1 - \sin x)^2} = \frac{\cos x}{(1 - \sin x)^2}$ . And $f(x)f'(x) = \frac{\cos x}{1 - \sin x}\cdot\frac{1}{1 - \sin x} = \frac{\cos x}{(1 - \sin x)^2} = f''(x)$ . At $0$ : $f''(0) = \frac{1}{1} = 1$ . Differentiating the simplified form of c) takes one line; differentiating $\sec x\tan x + \sec^2 x$ directly takes two product rules and an identity.

e) $f'(x) = 2 \iff \frac{1}{1 - \sin x} = 2 \iff 1 - \sin x = \frac{1}{2} \iff \sin x = \frac{1}{2}$ . On $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ the only solution is $x = \frac{\pi}{6}$ . There $f\left(\frac{\pi}{6}\right) = \frac{1 + 1/2}{\sqrt 3/2} = \frac{3}{\sqrt 3} = \sqrt 3$ . Tangent: $y = \sqrt 3 + 2\left(x - \frac{\pi}{6}\right) = 2x + \sqrt 3 - \frac{\pi}{3}$ , with $y$ -intercept $\sqrt 3 - \frac{\pi}{3} \approx 0.6849$ . The solution figure shows it, steeper than the tangent $y = 1 + x$ at the origin, as $f'' > 0$ on $\left(0, \frac{\pi}{6}\right)$ lets one expect.

Corrected exercises: derivatives of trigonometric functions (MATH 203)

Course recap

Part A: the basics (/50)

Exercise 1: The six derivatives, and the identity that comes BEFORE differentiating

Exercise 2: The product rule with trigonometric factors: name u and v, then collect

Exercise 3: The quotient rule, then the identity that comes AFTER

Exercise 4: Tangent and normal lines with exact trigonometric values

Exercise 5: Higher derivatives by the cycle of four, and the equation y'' + y = 0

Part B: problems and reasoning (/50)

Exercise 6: Horizontal and parallel tangents on an interval: one function, factor, every quadrant

Exercise 7: Reading f and f' on one figure: secant, tangent and their derivatives

Exercise 8: Five statements to correct

Exercise 9: A Ferris wheel: height, vertical velocity and the equation y'' + y = 0

Exercise 10: A final exam question: the function sec x + tan x

See also

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