Exercise 1: Where the derivative of sine comes from: the limit of sin t over t

Every derivative of this chapter rests on one limit, $\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1$ . Direct substitution gives $\frac{0}{0}$ , so the limit has to be PROVED, and the proof is geometric. This exercise builds it, then draws its first consequences.

The figure shows the unit circle, an angle $\theta$ measured in radians with $0 < \theta < \frac{\pi}{2}$ , the points $O(0, 0)$ , $A(1, 0)$ , $P(\cos\theta, \sin\theta)$ , its projection $Q(\cos\theta, 0)$ , and the point $T$ where the line $OP$ meets the tangent to the circle at $A$ , so that $AT = \tan\theta$ .

a) Compare the areas of the triangle $OAP$ , of the circular sector $OAP$ and of the triangle $OAT$ , and deduce that $\sin\theta < \theta < \tan\theta$ for $0 < \theta < \frac{\pi}{2}$ . Where exactly do radians enter?
b) Deduce that $\cos\theta < \frac{\sin\theta}{\theta} < 1$ on $\left(0, \frac{\pi}{2}\right)$ , then prove that $\lim_{\theta\to 0}\frac{\sin\theta}{\theta} = 1$ , from both sides.
c) Prove that $\lim_{\theta\to 0}\frac{1 - \cos\theta}{\theta} = 0$ and that $\lim_{\theta\to 0}\frac{1 - \cos\theta}{\theta^2} = \frac{1}{2}$ .
d) If $x$ is measured in degrees, $\sin(x^\circ) = \sin\left(\frac{\pi x}{180}\right)$ . Find $\lim_{x\to 0}\frac{\sin(x^\circ)}{x}$ , and explain why calculus measures angles in radians.
e) Find $\lim_{\theta\to 0}\frac{\tan\theta}{\theta}$ and $\lim_{\theta\to 0}\frac{\theta}{\sin\theta}$ , and say what the three lengths $PQ$ , arc $AP$ and $AT$ of the figure do as $\theta \to 0^+$ .

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Answers

a) $\frac{1}{2}\sin\theta < \frac{1}{2}\theta < \frac{1}{2}\tan\theta$ ; radians enter through the sector area $\frac{\theta}{2}$ .
b) Squeeze between $\cos\theta \to 1$ and $1$ ; the quotient is even, so both sides give $1$ .
c) $\frac{1 - \cos\theta}{\theta} = \frac{\sin\theta}{\theta}\cdot\frac{\sin\theta}{1 + \cos\theta} \to 0$ ; $\frac{1 - \cos\theta}{\theta^2} \to \frac{1}{2}$
d) $\frac{\pi}{180}$ , not $1$
e) $1$ and $1$ : the three lengths have ratios tending to $1$ .

a)

The triangle $OAP$ has base $OA = 1$ and height $PQ = \sin\theta$ , so its area is $\frac{1}{2}\sin\theta$ . The sector $OAP$ is the fraction $\frac{\theta}{2\pi}$ of the disc of area $\pi$ , so its area is $\frac{\theta}{2}$ . The triangle $OAT$ is right-angled at $A$ with legs $OA = 1$ and $AT = \tan\theta$ , so its area is $\frac{1}{2}\tan\theta$ . The first region lies strictly inside the second, which lies strictly inside the third, as the figure shows: $\frac{1}{2}\sin\theta < \frac{\theta}{2} < \frac{1}{2}\tan\theta$ , and multiplying by $2$ gives $\sin\theta < \theta < \tan\theta$ . Radians enter in ONE place, the sector area $\frac{\theta}{2}$ : with an angle of $d$ degrees the sector has area $\frac{\pi d}{360}$ , and the whole chain of the chapter would carry that factor.

b)

Dividing $\sin\theta < \theta$ by $\theta > 0$ gives $\frac{\sin\theta}{\theta} < 1$ . From $\theta < \tan\theta = \frac{\sin\theta}{\cos\theta}$ , multiplying by $\cos\theta > 0$ and dividing by $\theta > 0$ gives $\cos\theta < \frac{\sin\theta}{\theta}$ . So $\cos\theta < \frac{\sin\theta}{\theta} < 1$ on $\left(0, \frac{\pi}{2}\right)$ . As $\theta \to 0^+$ , $\cos\theta \to \cos 0 = 1$ because cosine is continuous, and the right bound is the constant $1$ : by the squeeze theorem, $\lim_{\theta\to 0^+}\frac{\sin\theta}{\theta} = 1$ . For $\theta < 0$ , $\frac{\sin(-\theta)}{-\theta} = \frac{-\sin\theta}{-\theta} = \frac{\sin\theta}{\theta}$ : the quotient is even, so the limit from the left equals the limit from the right. The two one-sided limits exist and are equal, hence the limit is $1$ . The quotient has no value at $0$ ; the limit never looks at that point.

c)

Multiply by the conjugate $1 + \cos\theta$ , which is close to $2$ near $0$ : $\frac{1 - \cos\theta}{\theta} = \frac{1 - \cos^2\theta}{\theta(1 + \cos\theta)} = \frac{\sin^2\theta}{\theta(1 + \cos\theta)} = \frac{\sin\theta}{\theta}\cdot\frac{\sin\theta}{1 + \cos\theta}$ . The first factor tends to $1$ by b), the second to $\frac{0}{2} = 0$ , so the product tends to $0$ . With $\theta^2$ in the denominator: $\frac{1 - \cos\theta}{\theta^2} = \left(\frac{\sin\theta}{\theta}\right)^2\cdot\frac{1}{1 + \cos\theta} \to 1 \cdot \frac{1}{2} = \frac{1}{2}$ . The two answers differ because $1 - \cos\theta$ shrinks like $\frac{\theta^2}{2}$ , much faster than $\theta$ : writing $1$ for the first limit, by analogy with sine, is the classic error.

d)

Put $u = \frac{\pi x}{180}$ , which tends to $0$ with $x$ . Then $\frac{\sin(x^\circ)}{x} = \frac{\sin u}{x} = \frac{\pi}{180}\cdot\frac{\sin u}{u} \to \frac{\pi}{180}\cdot 1 = \frac{\pi}{180}$ . In degrees the fundamental limit is $\frac{\pi}{180}$ , a little under $0.02$ , and every derivative of the chapter would be multiplied by it: the derivative of $\sin(x^\circ)$ would be $\frac{\pi}{180}\cos(x^\circ)$ . Radians are precisely the unit in which the limit equals $1$ , which is why every formula of calculus assumes them.

e)

$\frac{\tan\theta}{\theta} = \frac{\sin\theta}{\theta}\cdot\frac{1}{\cos\theta} \to 1 \cdot \frac{1}{1} = 1$ , by the product law and the continuity of cosine at $0$ . And $\frac{\theta}{\sin\theta} = \frac{1}{\sin\theta/\theta} \to \frac{1}{1} = 1$ , by the quotient law, legitimate because the limit of the denominator is $1 \ne 0$ . On the figure, $PQ = \sin\theta$ , arc $AP = \theta$ and $AT = \tan\theta$ : all three tend to $0$ , but their RATIOS tend to $1$ . For a small angle, the half-chord, the arc and the tangent segment are practically the same length, and this is the geometric content of the limit.

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Exercise 2: Limits built on sin(u)/u and 1 - cos(u): match the box before you cancel

The two limits of Exercise 1 are used in the forms $\lim_{\square\to 0}\frac{\sin\square}{\square} = 1$ and $\lim_{\square\to 0}\frac{1 - \cos\square}{\square^2} = \frac{1}{2}$ , where $\square$ is ANY expression that tends to $0$ , on the condition that the SAME expression sits in the numerator and in the denominator. Most of the work is to make the right box appear, by multiplying and dividing by a constant, by the conjugate or by factoring, before any limit law is used.

The figure shows $y = \frac{\sin x}{x}$ , with no value at $x = 0$ , between the dashed curves $y = \frac{1}{x}$ and $y = -\frac{1}{x}$ .

a) Find $\lim_{x\to 0}\frac{1 - \cos 4x}{x^2}$ and $\lim_{x\to 0}\frac{1 - \cos 3x}{1 - \cos x}$ .
b) Find $\lim_{x\to 0}\frac{\tan 4x}{x}$ and $\lim_{x\to 0} x\cot 2x$ .
c) Find $\lim_{x\to 0}\frac{1 - \cos x}{x\sin x}$ and $\lim_{x\to 0}\frac{\tan x - \sin x}{x^3}$ . For the second, a student replaces $\tan x$ and $\sin x$ by $x$ and answers $0$ : explain the error.
d) Recognize each limit as a derivative and compute it: $\lim_{x\to\pi/6}\frac{\sin x - \frac{1}{2}}{x - \frac{\pi}{6}}$ and $\lim_{h\to 0}\frac{\cos\left(\frac{\pi}{3} + h\right) - \frac{1}{2}}{h}$ .
e) Find $\lim_{x\to\infty}\frac{\sin x}{x}$ and $\lim_{x\to\infty} x\sin\frac{1}{x}$ . A student answers $1$ to both: which one is right, and why?

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Answers

a) $8$ and $9$
b) $4$ and $\frac{1}{2}$
c) $\frac{1}{2}$ and $\frac{1}{2}$ (not $0$ )
d) $\cos\frac{\pi}{6} = \frac{\sqrt 3}{2}$ and $-\sin\frac{\pi}{3} = -\frac{\sqrt 3}{2}$
e) $0$ and $1$ : only the second has a box $\frac{1}{x}$ that tends to $0$ .

a)

The box of $1 - \cos 4x$ is $4x$ , so $(4x)^2 = 16x^2$ must appear below: $\frac{1 - \cos 4x}{x^2} = 16\cdot\frac{1 - \cos 4x}{(4x)^2} \to 16 \cdot \frac{1}{2} = 8$ . Answering $\frac{1}{2}$ , or $4 \cdot \frac{1}{2} = 2$ , forgets that the box is SQUARED in the denominator. For the quotient of two cosines, divide top and bottom by $x^2$ : $\frac{1 - \cos 3x}{1 - \cos x} = \frac{9\cdot\frac{1 - \cos 3x}{(3x)^2}}{\frac{1 - \cos x}{x^2}} \to \frac{9 \cdot \frac{1}{2}}{\frac{1}{2}} = 9$ , the quotient law being legitimate because the limit of the denominator is $\frac{1}{2} \ne 0$ . The answer is $3^2$ , not $3$ : near $0$ , $1 - \cos u$ behaves like $\frac{u^2}{2}$ .

b)

A tangent is a sine over a cosine: $\frac{\tan 4x}{x} = \frac{\sin 4x}{x}\cdot\frac{1}{\cos 4x} = 4\cdot\frac{\sin 4x}{4x}\cdot\frac{1}{\cos 4x} \to 4 \cdot 1 \cdot \frac{1}{1} = 4$ . For the cotangent, $x\cot 2x = \frac{x\cos 2x}{\sin 2x} = \frac{1}{2}\cdot\frac{2x}{\sin 2x}\cdot\cos 2x \to \frac{1}{2}\cdot 1 \cdot 1 = \frac{1}{2}$ . The factor $\frac{2x}{\sin 2x}$ is the reciprocal of the standard quotient, and its limit is $\frac{1}{1} = 1$ by the quotient law. In both cases the $\cos$ that appears tends to $1$ and does no harm.

c)

Direct substitution gives $\frac{0}{0}$ . The numerator $1 - \cos x$ calls for the conjugate: $\frac{1 - \cos x}{x\sin x} = \frac{1 - \cos^2 x}{x\sin x\,(1 + \cos x)} = \frac{\sin x}{x}\cdot\frac{1}{1 + \cos x} \to \frac{1}{2}$ , cancelling one $\sin x$ , which is legitimate because $\sin x \ne 0$ for $0 < |x| < \pi$ . For the second, factor instead of approximating: $\tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \frac{\sin x(1 - \cos x)}{\cos x}$ , so $\frac{\tan x - \sin x}{x^3} = \frac{\sin x}{x}\cdot\frac{1 - \cos x}{x^2}\cdot\frac{1}{\cos x} \to 1 \cdot \frac{1}{2}\cdot 1 = \frac{1}{2}$ . The student's $0$ comes from replacing two terms of a DIFFERENCE by their first approximation $x$ : the approximations cancel exactly, and what is left, of order $x^3$ , is precisely what the question measures. Limit laws act on factors of a product, never on pieces of a difference.

d)

By definition, $f'(a) = \lim_{x\to a}\frac{f(x) - f(a)}{x - a} = \lim_{h\to 0}\frac{f(a + h) - f(a)}{h}$ . The first limit has the first form with $f = \sin$ and $a = \frac{\pi}{6}$ , since $\sin\frac{\pi}{6} = \frac{1}{2}$ : it is $\cos\frac{\pi}{6} = \frac{\sqrt 3}{2}$ . The second has the second form with $f = \cos$ and $a = \frac{\pi}{3}$ , since $\cos\frac{\pi}{3} = \frac{1}{2}$ : it is $-\sin\frac{\pi}{3} = -\frac{\sqrt 3}{2}$ . Recognizing $f$ AND $a$ is the whole question; the only trap is the sign of $(\cos x)'$ . Computing these limits directly would mean redoing the proof of Exercise 3 at a particular point: the derivative is the shortcut that the proof has earned.

e)

In $\frac{\sin x}{x}$ the box $x$ does NOT tend to $0$ , and the standard limit says nothing. Use the squeeze theorem instead: for $x > 0$ , $-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$ because $|\sin x| \le 1$ , and both bounds tend to $0$ , so the limit is $0$ ; the figure shows the curve pinned between the dashed envelopes. In $x\sin\frac{1}{x}$ , on the contrary, write $u = \frac{1}{x}$ , which tends to $0^+$ as $x \to \infty$ : $x\sin\frac{1}{x} = \frac{\sin u}{u} \to 1$ . Same variable, same destination, two different answers: what decides is whether the quantity under the sine tends to $0$ , never the look of the expression. The student is right about the second only.

Exercise 3: The six derivatives: two proved from the definition, four by the quotient rule

By definition $f'(x) = \lim_{h\to 0}\frac{f(x + h) - f(x)}{h}$ . For sine and cosine the difference $f(x + h) - f(x)$ is opened with the addition formulas $\sin(x + h) = \sin x\cos h + \cos x\sin h$ and $\cos(x + h) = \cos x\cos h - \sin x\sin h$ ; the two limits of Exercise 1 then do the rest. The four other functions are quotients of these two.

The figure shows $y = \sin x$ and three of its tangent lines, at $x = 0$ , $x = \frac{\pi}{2}$ and $x = \pi$ .

a) Prove from the definition that $(\sin x)' = \cos x$ , naming each limit used.
b) Prove from the definition that $(\cos x)' = -\sin x$ .
c) Give the slopes and the equations of the three tangent lines of the figure, and check that the drawing agrees.
d) Using the quotient rule, prove that $(\tan x)' = \sec^2 x$ and $(\sec x)' = \sec x\tan x$ , and say where these formulas hold.
e) Prove that $(\cot x)' = -\csc^2 x$ and $(\csc x)' = -\csc x\cot x$ . State the rule that gives the signs of all six derivatives.

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Answers

a) $\frac{\sin(x + h) - \sin x}{h} = \sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h} \to \cos x$
b) $\frac{\cos(x + h) - \cos x}{h} = \cos x\cdot\frac{\cos h - 1}{h} - \sin x\cdot\frac{\sin h}{h} \to -\sin x$
c) Slopes $1$ , $0$ , $-1$ : $y = x$ , $y = 1$ , $y = \pi - x$
d) $(\tan x)' = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \sec^2 x$ , $(\sec x)' = \frac{\sin x}{\cos^2 x} = \sec x\tan x$ , for $\cos x \ne 0$
e) $(\cot x)' = -\csc^2 x$ , $(\csc x)' = -\csc x\cot x$ , for $\sin x \ne 0$ ; the three co-functions take a minus sign.

a)

With $f(x) = \sin x$ and the addition formula, $\frac{\sin(x + h) - \sin x}{h} = \frac{\sin x\cos h + \cos x\sin h - \sin x}{h} = \sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}$ . Here $x$ is FIXED and only $h$ moves, so $\sin x$ and $\cos x$ are constants for the limit. By Exercise 1, $\frac{\cos h - 1}{h} \to 0$ and $\frac{\sin h}{h} \to 1$ as $h \to 0$ , with $h$ in radians. By the sum and product laws, the limit is $\sin x \cdot 0 + \cos x \cdot 1 = \cos x$ . The regrouping by $\sin x$ and $\cos x$ is the whole trick: it turns one unknown limit into two known ones.

b)

$\frac{\cos(x + h) - \cos x}{h} = \frac{\cos x\cos h - \sin x\sin h - \cos x}{h} = \cos x\cdot\frac{\cos h - 1}{h} - \sin x\cdot\frac{\sin h}{h} \to \cos x\cdot 0 - \sin x\cdot 1 = -\sin x$ . The minus sign comes from the addition formula of cosine, and nowhere else: a student who remembers the formula remembers the sign.

c)

The slope of the tangent to $y = \sin x$ at $a$ is $\cos a$ . At $0$ : $\cos 0 = 1$ , through $(0, 0)$ , so $y = x$ . At $\frac{\pi}{2}$ : $\cos\frac{\pi}{2} = 0$ , through $\left(\frac{\pi}{2}, 1\right)$ , so the horizontal line $y = 1$ . At $\pi$ : $\cos\pi = -1$ , through $(\pi, 0)$ , so $y = -(x - \pi) = \pi - x$ . The figure agrees: the first tangent passes through $(1, 1)$ , one unit up for one unit across; the second is flat on the top of the arch; the third crosses the axis at $\pi$ and passes through $(\pi - 1, 1)$ . Reading slopes off a graph is the fastest way to catch a sign error in $(\cos x)'$ .

d)

Wherever $\cos x \ne 0$ , $\tan x = \frac{\sin x}{\cos x}$ , and the quotient rule gives $(\tan x)' = \frac{\cos x\cdot\cos x - \sin x\cdot(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$ , using $\sin^2 x + \cos^2 x = 1$ ; equivalently $1 + \tan^2 x$ . Next, $\sec x = \frac{1}{\cos x}$ : $(\sec x)' = \frac{0\cdot\cos x - 1\cdot(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x\tan x$ . Both hold exactly where the functions are defined, $x \ne \frac{\pi}{2} + k\pi$ for every integer $k$ . The minus of $(\cos x)'$ meets the minus of the quotient rule, and the two cancel: that is why these two derivatives are positive-looking.

e)

Wherever $\sin x \ne 0$ : $(\cot x)' = \left(\frac{\cos x}{\sin x}\right)' = \frac{(-\sin x)\sin x - \cos x\cos x}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = -\csc^2 x$ , and $(\csc x)' = \left(\frac{1}{\sin x}\right)' = \frac{0 - \cos x}{\sin^2 x} = -\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = -\csc x\cot x$ , for $x \ne k\pi$ . The rule of signs: the three CO-functions, $\cos$ , $\cot$ and $\csc$ , have a minus sign in their derivatives, and each co-derivative is obtained from its partner by swapping every function with its co-function and adding the minus: $\tan \to \sec^2$ becomes $\cot \to -\csc^2$ , and $\sec \to \sec\tan$ becomes $\csc \to -\csc\cot$ .

Exercise 4: Product and quotient rules with trigonometric functions, then simplify

Once the six derivatives are known, a trigonometric expression is differentiated with the rules of the previous chapter: sum, product $(uv)' = u'v + uv'$ , quotient $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ . The derivative of a product is NOT the product of the derivatives, and the derivative of a quotient is NOT the quotient of the derivatives.

A trigonometric derivative is rarely finished when the rule has been applied: the identities $\sin^2 x + \cos^2 x = 1$ and $\sec^2 x = 1 + \tan^2 x$ usually collapse the numerator, and the exam expects the simplified form. The chain rule is the next chapter: it is not used here.

a) Differentiate $f(x) = x^2\sin x$ and $g(x) = e^x\cos x$ .
b) Differentiate $h(x) = \frac{\sin x}{1 + \cos x}$ and simplify the result to a single short fraction.
c) Differentiate $k(x) = \frac{\sec x}{1 + \tan x}$ , simplify, and find $k'\left(\frac{\pi}{4}\right)$ .
d) Differentiate $y = \sin x\cos x$ . Deduce the derivative of $\sin 2x$ WITHOUT the chain rule, using $\sin 2x = 2\sin x\cos x$ and $\cos 2x = \cos^2 x - \sin^2 x$ .
e) Differentiate $p(x) = \tan x\cos x$ in two ways: by the product rule, then after simplifying $p$ . Compare, and state the domain of $p$ and of $p'$ .

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Answers

a) $f'(x) = 2x\sin x + x^2\cos x$ ; $g'(x) = e^x(\cos x - \sin x)$
b) $h'(x) = \frac{1}{1 + \cos x}$
c) $k'(x) = \frac{\sec x(\tan x - 1)}{(1 + \tan x)^2}$ ; $k'\left(\frac{\pi}{4}\right) = 0$
d) $y' = \cos^2 x - \sin^2 x$ ; $(\sin 2x)' = 2\cos 2x$
e) Both give $p'(x) = \cos x$ , but only for $\cos x \ne 0$ : $p$ and $p'$ are undefined at $\frac{\pi}{2} + k\pi$ .

a)

Product rule with $u = x^2$ , $v = \sin x$ , $u' = 2x$ , $v' = \cos x$ : $f'(x) = 2x\sin x + x^2\cos x$ . Product rule with $u = e^x$ , $v = \cos x$ , $u' = e^x$ , $v' = -\sin x$ : $g'(x) = e^x\cos x - e^x\sin x = e^x(\cos x - \sin x)$ . Naming $u$ , $v$ , $u'$ and $v'$ on the copy before assembling is what makes the minus of $(\cos x)'$ impossible to lose. Writing $(x^2\sin x)' = 2x\cos x$ , the product of the derivatives, is worth zero.

b)

Quotient rule with $u = \sin x$ , $v = 1 + \cos x$ , $u' = \cos x$ , $v' = -\sin x$ : $h'(x) = \frac{\cos x(1 + \cos x) - \sin x(-\sin x)}{(1 + \cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2} = \frac{1 + \cos x}{(1 + \cos x)^2} = \frac{1}{1 + \cos x}$ , wherever $\cos x \ne -1$ . The two minus signs, from $v'$ and from the quotient rule, combine into a plus, and then $\cos^2 x + \sin^2 x = 1$ produces the factor $1 + \cos x$ that cancels. Stopping at the unsimplified fraction usually costs a mark on a question that says simplify.

c)

Here $u = \sec x$ , $u' = \sec x\tan x$ , $v = 1 + \tan x$ , $v' = \sec^2 x$ . So $k'(x) = \frac{\sec x\tan x(1 + \tan x) - \sec x\cdot\sec^2 x}{(1 + \tan x)^2} = \frac{\sec x\left(\tan x + \tan^2 x - \sec^2 x\right)}{(1 + \tan x)^2}$ . With $\sec^2 x = 1 + \tan^2 x$ the bracket becomes $\tan x - 1$ : $k'(x) = \frac{\sec x(\tan x - 1)}{(1 + \tan x)^2}$ , valid where $\cos x \ne 0$ and $\tan x \ne -1$ . At $\frac{\pi}{4}$ , $\tan\frac{\pi}{4} = 1$ , so $k'\left(\frac{\pi}{4}\right) = 0$ : the graph of $k$ has a horizontal tangent there. Factoring $\sec x$ out first is what lets the identity act.

d)

Product rule: $y' = (\sin x)'\cos x + \sin x(\cos x)' = \cos x\cos x + \sin x(-\sin x) = \cos^2 x - \sin^2 x$ . Since $\sin 2x = 2\sin x\cos x = 2y$ , the constant multiple rule gives $(\sin 2x)' = 2(\cos^2 x - \sin^2 x) = 2\cos 2x$ . The factor $2$ is the one students drop when they write $(\sin 2x)' = \cos 2x$ by analogy with $(\sin x)' = \cos x$ : the formula $(\sin x)' = \cos x$ is about the argument $x$ alone. Next chapter, the chain rule will produce the same $2$ mechanically; here the identity gives it for free, and checks it.

e)

Product rule with $u = \tan x$ , $v = \cos x$ : $p'(x) = \sec^2 x\cos x + \tan x(-\sin x) = \frac{1}{\cos x} - \frac{\sin^2 x}{\cos x} = \frac{1 - \sin^2 x}{\cos x} = \frac{\cos^2 x}{\cos x} = \cos x$ . After simplifying first, $p(x) = \frac{\sin x}{\cos x}\cos x = \sin x$ wherever $\cos x \ne 0$ , so $p'(x) = \cos x$ at once. Same answer, far less work: simplify BEFORE differentiating whenever you can. But the domain does not simplify: $p$ is undefined at $x = \frac{\pi}{2} + k\pi$ , its graph is the sine curve with holes there, and $p'$ does not exist at those points either, although the formula $\cos x$ has a value there. The correct answer is $p'(x) = \cos x$ for $\cos x \ne 0$ .

Exercise 5: Derivatives of high order: the cycle of four

Differentiating sine four times brings it back: the derivatives of $\sin x$ and $\cos x$ repeat with period $4$ . A derivative of order $99$ is therefore read off from a remainder, never computed ninety-nine times. The figure shows $y = \sin x$ and $y = \cos x$ : each derivative shifts the graph of sine to the LEFT by a quarter period, $\frac{\pi}{2}$ .

a) Write the first four derivatives of $\sin x$ and of $\cos x$ , and state how $\frac{d^n}{dx^n}\sin x$ depends on $n$ .
b) Find $\frac{d^{99}}{dx^{99}}\sin x$ and $\frac{d^{50}}{dx^{50}}\cos x$ .
c) Let $y = x\cos x$ . Compute $y'$ , $y''$ , $y'''$ and $y^{(4)}$ , then prove by induction that $y^{(n)} = x\,c^{(n)}(x) + n\,c^{(n-1)}(x)$ for $n \ge 1$ , where $c = \cos$ . Deduce $y^{(42)}$ .
d) Find $(\tan x)''$ and $(\sec x)''$ , writing $\sec^2 x$ as the PRODUCT $\sec x\cdot\sec x$ . Show that $(\sec x)'' = 2\sec^3 x - \sec x$ .
e) Let $f(x) = \sin x + \cos x$ . Find $f^{(n)}(0)$ for every $n \ge 0$ , and count the $n$ between $1$ and $100$ for which $f^{(n)}(0) = -1$ .

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Answers

a) $\sin \to \cos \to -\sin \to -\cos \to \sin$ ; the $n$ -th derivative depends only on the remainder of $n$ divided by $4$ .
b) $-\cos x$ and $-\cos x$
c) $y^{(42)} = -x\cos x - 42\sin x$
d) $(\tan x)'' = 2\sec^2 x\tan x$ ; $(\sec x)'' = \sec x\tan^2 x + \sec^3 x = 2\sec^3 x - \sec x$
e) $1, 1, -1, -1$ repeating; $f^{(n)}(0) = -1$ for $50$ values of $n$ .

a)

$(\sin x)' = \cos x$ , $(\sin x)'' = -\sin x$ , $(\sin x)''' = -\cos x$ , $(\sin x)^{(4)} = \sin x$ . And $(\cos x)' = -\sin x$ , $(\cos x)'' = -\cos x$ , $(\cos x)''' = \sin x$ , $(\cos x)^{(4)} = \cos x$ . Since the fourth derivative is the function itself, $\frac{d^{4k + r}}{dx^{4k + r}}\sin x = \frac{d^r}{dx^r}\sin x$ for $r = 0, 1, 2, 3$ : only the remainder $r$ of $n$ divided by $4$ matters. On the figure, the peak of $\cos x$ at $0$ is the peak of $\sin x$ at $\frac{\pi}{2}$ moved left by $\frac{\pi}{2}$ , and indeed $\cos x = \sin\left(x + \frac{\pi}{2}\right)$ : four quarter-shifts make a full period.

b)

$99 = 4 \times 24 + 3$ , so $\frac{d^{99}}{dx^{99}}\sin x = (\sin x)''' = -\cos x$ . $50 = 4 \times 12 + 2$ , so $\frac{d^{50}}{dx^{50}}\cos x = (\cos x)'' = -\cos x$ . The common error is to reason on parity alone, 99 is odd so the answer is $\cos x$ : parity decides between sine and cosine, but the SIGN needs the remainder modulo $4$ . Check the remainder by multiplying back: $4 \times 24 = 96$ and $96 + 3 = 99$ .

c)

$y' = \cos x - x\sin x$ (product rule). $y'' = -\sin x - \sin x - x\cos x = -2\sin x - x\cos x$ . $y''' = -2\cos x - \cos x + x\sin x = -3\cos x + x\sin x$ . $y^{(4)} = 3\sin x + \sin x + x\cos x = 4\sin x + x\cos x$ . Each line matches $x\,c^{(n)}(x) + n\,c^{(n-1)}(x)$ , since $c' = -\sin$ , $c'' = -\cos$ , $c''' = \sin$ , $c^{(4)} = \cos$ . Induction: the formula holds for $n = 1$ ; if $y^{(n)} = x\,c^{(n)} + n\,c^{(n-1)}$ , the product rule gives $y^{(n+1)} = c^{(n)} + x\,c^{(n+1)} + n\,c^{(n)} = x\,c^{(n+1)} + (n + 1)\,c^{(n)}$ , the formula for $n + 1$ . For $n = 42$ : $42 = 4 \times 10 + 2$ gives $c^{(42)} = -\cos x$ , and $41 = 4 \times 10 + 1$ gives $c^{(41)} = -\sin x$ . So $y^{(42)} = -x\cos x - 42\sin x$ . The coefficient $42$ counts how many times the factor $x$ has been differentiated away, once in each of the $42$ steps.

d)

$(\tan x)' = \sec^2 x = \sec x\cdot\sec x$ , and the product rule gives $(\tan x)'' = \sec x\tan x\cdot\sec x + \sec x\cdot\sec x\tan x = 2\sec^2 x\tan x$ . Writing $\sec^2 x$ as a product is what makes this possible before the chain rule. Next, $(\sec x)'' = (\sec x\tan x)' = \sec x\tan x\cdot\tan x + \sec x\cdot\sec^2 x = \sec x\tan^2 x + \sec^3 x$ . With $\tan^2 x = \sec^2 x - 1$ : $\sec x(\sec^2 x - 1) + \sec^3 x = 2\sec^3 x - \sec x$ . Check at $x = 0$ : the first form gives $\sec 0\tan^2 0 + \sec^3 0 = 0 + 1 = 1$ and the second $2 - 1 = 1$ . Two forms that disagree at one convenient point cannot both be right.

e)

$f(0) = 1$ . $f' = \cos x - \sin x$ , so $f'(0) = 1$ . $f'' = -\sin x - \cos x$ , so $f''(0) = -1$ . $f''' = -\cos x + \sin x$ , so $f'''(0) = -1$ . $f^{(4)} = f$ , so the values repeat: $1, 1, -1, -1, 1, 1, -1, -1, \dots$ Hence $f^{(n)}(0) = 1$ when $n$ leaves remainder $0$ or $1$ on division by $4$ , and $-1$ when it leaves $2$ or $3$ . Between $1$ and $100$ there are $25$ complete blocks of four consecutive integers, each containing exactly two values with remainder $2$ or $3$ : $50$ values of $n$ .

Exercise 6: Horizontal tangents on an interval: factor, never divide

A curve $y = f(x)$ has a horizontal tangent at $a$ when $f'(a) = 0$ . For a trigonometric function the equation $f'(x) = 0$ is a trigonometric equation, solved on the interval asked, with EXACT solutions. Its classic trap: dividing both sides by $\sin x$ or $\cos x$ silently throws away the solutions where that factor vanishes. Factor, then set each factor to zero.

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

a) Find the points of $f(x) = x + 2\sin x$ , $0 \le x \le 2\pi$ , where the tangent is horizontal.
b) Same question for $g(x) = \cos x + \sin^2 x$ , writing $\sin^2 x = \sin x\cdot\sin x$ . How many points does the figure show, and how many does a student find who divides by $\sin x$ ?
c) Same question for $h(x) = e^x\sin x$ on $[0, 2\pi]$ . Here dividing by $\cos x$ is legitimate: prove it.
d) Find the horizontal tangents of $y = \sec x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and of $y = \csc x$ on $(0, \pi)$ . Does $y = \tan x$ have any?
e) For which real constants $c$ does $y = x + c\sin x$ have at least one horizontal tangent? Give the points when $c = 1$ .

Type your answers, the page tells you right or wrong 0/12

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Show the solution

Answers

a) $\left(\frac{2\pi}{3}, \frac{2\pi}{3} + \sqrt 3\right)$ and $\left(\frac{4\pi}{3}, \frac{4\pi}{3} - \sqrt 3\right)$
b) Five points: $(0, 1)$ , $\left(\frac{\pi}{3}, \frac{5}{4}\right)$ , $(\pi, -1)$ , $\left(\frac{5\pi}{3}, \frac{5}{4}\right)$ , $(2\pi, 1)$ ; dividing keeps only two.
c) $x = \frac{3\pi}{4}$ and $x = \frac{7\pi}{4}$ : $\left(\frac{3\pi}{4}, \frac{\sqrt 2}{2}e^{3\pi/4}\right)$ , $\left(\frac{7\pi}{4}, -\frac{\sqrt 2}{2}e^{7\pi/4}\right)$
d) $\sec$ : $(0, 1)$ ; $\csc$ : $\left(\frac{\pi}{2}, 1\right)$ ; $\tan$ : none, since $\sec^2 x \ge 1$ .
e) $|c| \ge 1$ ; for $c = 1$ , the points $(\pi + 2k\pi, \pi + 2k\pi)$ , $k$ an integer.

a)

$f'(x) = 1 + 2\cos x$ , by the sum rule and $(\sin x)' = \cos x$ . Then $f'(x) = 0 \iff \cos x = -\frac{1}{2}$ . On $[0, 2\pi]$ , cosine equals $-\frac{1}{2}$ at $x = \frac{2\pi}{3}$ (second quadrant) and $x = \frac{4\pi}{3}$ (third quadrant), and nowhere else. The heights: $f\left(\frac{2\pi}{3}\right) = \frac{2\pi}{3} + 2\cdot\frac{\sqrt 3}{2} = \frac{2\pi}{3} + \sqrt 3$ and $f\left(\frac{4\pi}{3}\right) = \frac{4\pi}{3} - \sqrt 3$ . A point is two coordinates: giving only the $x$ values answers half the question.

b)

$g'(x) = -\sin x + (\sin x\cdot\sin x)' = -\sin x + 2\sin x\cos x$ , the product rule giving $\cos x\sin x + \sin x\cos x$ . Factor: $g'(x) = \sin x\,(2\cos x - 1)$ . A product is zero when one of its factors is: $\sin x = 0$ gives $x = 0, \pi, 2\pi$ ; $\cos x = \frac{1}{2}$ gives $x = \frac{\pi}{3}, \frac{5\pi}{3}$ . Heights: $g(0) = 1$ , $g\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{3}{4} = \frac{5}{4}$ , $g(\pi) = -1 + 0 = -1$ , $g\left(\frac{5\pi}{3}\right) = \frac{5}{4}$ , $g(2\pi) = 1$ . The figure shows exactly these five flat spots: the two bumps, the bottom of the valley at $\pi$ , and the two ends. The student who writes $-\sin x + 2\sin x\cos x = 0 \Rightarrow 2\cos x = 1$ divided by $\sin x$ , which is legitimate only where $\sin x \ne 0$ , and lost three of the five points without noticing.

c)

$h'(x) = e^x\sin x + e^x\cos x = e^x(\sin x + \cos x)$ by the product rule. Since $e^x > 0$ , $h'(x) = 0 \iff \sin x + \cos x = 0$ . Before dividing by $\cos x$ , check that no solution is lost: if $\cos x = 0$ then $\sin x = \pm 1$ and $\sin x + \cos x = \pm 1 \ne 0$ , so every solution has $\cos x \ne 0$ . Dividing is now safe: $\tan x = -1$ , so $x = \frac{3\pi}{4}$ or $x = \frac{7\pi}{4}$ on $[0, 2\pi]$ . Heights: $h\left(\frac{3\pi}{4}\right) = e^{3\pi/4}\cdot\frac{\sqrt 2}{2}$ and $h\left(\frac{7\pi}{4}\right) = -e^{7\pi/4}\cdot\frac{\sqrt 2}{2}$ , exact values, no decimal needed. The one-line check is the difference between a division that is a proof and a division that is a gamble.

d)

$(\sec x)' = \sec x\tan x$ . The factor $\sec x = \frac{1}{\cos x}$ is never $0$ , so $(\sec x)' = 0 \iff \tan x = 0 \iff x = 0$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ : the point $(0, 1)$ . $(\csc x)' = -\csc x\cot x$ , and $\csc x$ is never $0$ , so $\cot x = 0$ , that is $\cos x = 0$ with $\sin x \ne 0$ : $x = \frac{\pi}{2}$ , the point $\left(\frac{\pi}{2}, 1\right)$ . For the tangent function, $(\tan x)' = \sec^2 x = \frac{1}{\cos^2 x} \ge 1$ wherever it is defined: the slope is never $0$ , it is never even below $1$ , so $y = \tan x$ has no horizontal tangent at all.

e)

$y' = 1 + c\cos x$ . If $c = 0$ , $y' = 1$ and there is no horizontal tangent. Otherwise $y' = 0 \iff \cos x = -\frac{1}{c}$ , which has a solution exactly when $\left|-\frac{1}{c}\right| \le 1$ , that is $|c| \ge 1$ . Conversely, when $|c| < 1$ , $|c\cos x| \le |c| < 1$ , so $y' = 1 + c\cos x > 0$ for every $x$ . Answer: $|c| \ge 1$ , and $c = 2$ was question a). For $c = 1$ : $\cos x = -1$ , so $x = \pi + 2k\pi$ with $k$ an integer, where $\sin x = 0$ and $y = x$ : the points $(\pi + 2k\pi, \pi + 2k\pi)$ , infinitely many, all on the line $y = x$ .

Exercise 7: Tangent and normal lines to trigonometric curves

The tangent to $y = f(x)$ at $a$ is $y = f(a) + f'(a)(x - a)$ ; the normal is the line through the same point perpendicular to it, with slope $-\frac{1}{f'(a)}$ when $f'(a) \ne 0$ . Every value is exact: $\sec\frac{\pi}{3} = 2$ , $\tan\frac{\pi}{4} = 1$ , and $\pi$ stays $\pi$ in the final equation.

The figure, drawn with EQUAL scales on both axes so that angles are true, shows $y = \sin x$ and $y = \cos x$ crossing at the point $P$ of abscissa $\frac{\pi}{4}$ .

a) Find the tangent to $y = \tan x$ at $x = \frac{\pi}{4}$ .
b) Find the tangent and the normal to $y = x\cos x$ at $x = \pi$ .
c) Find the tangent to $y = 2\sec x - \cos x$ at $x = \frac{\pi}{3}$ .
d) A student claims that the curves $y = \sin x$ and $y = \cos x$ cross at right angles at $P$ . Decide, and give the angle between their tangent lines at $P$ in exact form.
e) Find every $x$ in $[0, 2\pi]$ where the tangent to $y = \sin x$ is parallel to the line $y = \frac{x}{2}$ , and write each tangent.

Show the solution

Answers

a) $y = 2x + 1 - \frac{\pi}{2}$
b) Tangent $y = -x$ ; normal $y = x - 2\pi$
c) $y = \frac{7}{2} + \frac{9\sqrt 3}{2}\left(x - \frac{\pi}{3}\right)$
d) False: slopes $\frac{\sqrt 2}{2}$ and $-\frac{\sqrt 2}{2}$ , product $-\frac{1}{2} \ne -1$ ; angle $\arctan(2\sqrt 2)$ .
e) $x = \frac{\pi}{3}$ : $y = \frac{\sqrt 3}{2} + \frac{1}{2}\left(x - \frac{\pi}{3}\right)$ ; $x = \frac{5\pi}{3}$ : $y = -\frac{\sqrt 3}{2} + \frac{1}{2}\left(x - \frac{5\pi}{3}\right)$

a)

Point: $\tan\frac{\pi}{4} = 1$ . Slope: $(\tan x)' = \sec^2 x$ and $\sec\frac{\pi}{4} = \frac{1}{\cos(\pi/4)} = \frac{2}{\sqrt 2} = \sqrt 2$ , so the slope is $(\sqrt 2)^2 = 2$ . Tangent: $y = 1 + 2\left(x - \frac{\pi}{4}\right) = 2x + 1 - \frac{\pi}{2}$ . Leaving $\frac{\pi}{2}$ exact is the expected form; replacing it by $1.57$ turns an exact answer into an approximate one on an exam without calculators.

b)

$y' = \cos x - x\sin x$ by the product rule. At $\pi$ : $y(\pi) = \pi\cos\pi = -\pi$ and $y'(\pi) = \cos\pi - \pi\sin\pi = -1 - 0 = -1$ . Tangent: $y = -\pi - (x - \pi) = -x$ , which passes through the origin. The normal has slope $-\frac{1}{-1} = 1$ : $y = -\pi + (x - \pi) = x - 2\pi$ . The slope of the normal is the NEGATIVE reciprocal; the reciprocal alone gives $-1$ again, a line parallel to the tangent.

c)

At $\frac{\pi}{3}$ : $\cos\frac{\pi}{3} = \frac{1}{2}$ , $\sec\frac{\pi}{3} = 2$ , $\sin\frac{\pi}{3} = \frac{\sqrt 3}{2}$ , $\tan\frac{\pi}{3} = \sqrt 3$ . Point: $y = 2 \cdot 2 - \frac{1}{2} = \frac{7}{2}$ . Derivative: $y' = 2\sec x\tan x - (-\sin x) = 2\sec x\tan x + \sin x$ , and the minus of $(\cos x)'$ meets the minus in front of $\cos x$ . Slope: $2 \cdot 2 \cdot \sqrt 3 + \frac{\sqrt 3}{2} = 4\sqrt 3 + \frac{\sqrt 3}{2} = \frac{9\sqrt 3}{2}$ . Tangent: $y = \frac{7}{2} + \frac{9\sqrt 3}{2}\left(x - \frac{\pi}{3}\right)$ . Writing $y' = 2\sec x\tan x - \sin x$ is the single most frequent sign slip of the chapter.

d)

At $x = \frac{\pi}{4}$ , $\sin x = \cos x = \frac{\sqrt 2}{2}$ , so $P\left(\frac{\pi}{4}, \frac{\sqrt 2}{2}\right)$ is on both curves. Slopes: $m_1 = \cos\frac{\pi}{4} = \frac{\sqrt 2}{2}$ for sine, $m_2 = -\sin\frac{\pi}{4} = -\frac{\sqrt 2}{2}$ for cosine. Two lines are perpendicular exactly when $m_1 m_2 = -1$ ; here $m_1 m_2 = -\frac{2}{4} = -\frac{1}{2}$ . The claim is FALSE: the curves cross at an angle smaller than a right angle, as the figure, drawn at equal scales, shows. The angle $\varphi$ between the tangents satisfies $\tan\varphi = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \frac{\sqrt 2}{1/2} = 2\sqrt 2$ , so $\varphi = \arctan(2\sqrt 2)$ . Equivalently each tangent makes the angle $\arctan\frac{\sqrt 2}{2}$ with the horizontal, on opposite sides, and $\tan\left(2\arctan\frac{\sqrt 2}{2}\right) = \frac{\sqrt 2}{1 - \frac{1}{2}} = 2\sqrt 2$ confirms it. The symmetry of the picture suggests a right angle; the slopes refute it.

e)

Parallel means equal slopes: $\cos x = \frac{1}{2}$ , so $x = \frac{\pi}{3}$ or $x = \frac{5\pi}{3}$ on $[0, 2\pi]$ . At $\frac{\pi}{3}$ the point is $\left(\frac{\pi}{3}, \frac{\sqrt 3}{2}\right)$ and the tangent is $y = \frac{\sqrt 3}{2} + \frac{1}{2}\left(x - \frac{\pi}{3}\right)$ . At $\frac{5\pi}{3}$ , $\sin\frac{5\pi}{3} = -\frac{\sqrt 3}{2}$ and the tangent is $y = -\frac{\sqrt 3}{2} + \frac{1}{2}\left(x - \frac{5\pi}{3}\right)$ . The two tangents are parallel to each other and distinct: at $x = 0$ they give $\frac{\sqrt 3}{2} - \frac{\pi}{6}$ and $-\frac{\sqrt 3}{2} - \frac{5\pi}{6}$ .

Exercise 8: Five statements to correct

Each statement below was written by a student on a MATH 140 assignment, and each is false. Say what is wrong, support the correction with a counterexample or a computation, and write the correct statement.

a) The formula $(\sin x)' = \cos x$ holds whether $x$ is measured in radians or in degrees.
b) $\lim_{x\to 0}\frac{1 - \cos 2x}{x^2} = \frac{1}{2}$ , because $\frac{1 - \cos\square}{\square^2} \to \frac{1}{2}$ .
c) $(\csc x)' = \csc x\cot x$ , just as $(\sec x)' = \sec x\tan x$ .
d) $\lim_{x\to 0}\frac{1 - \cos x}{x} = 1$ , just as $\lim_{x\to 0}\frac{\sin x}{x} = 1$ .
e) $(\tan x)' = \frac{(\sin x)'}{(\cos x)'} = \frac{\cos x}{-\sin x} = -\cot x$ .

Show the solution

Answers

a) False: in degrees, $\frac{d}{dx}\sin(x^\circ) = \frac{\pi}{180}\cos(x^\circ)$ .
b) False: the box is $2x$ , squared below; the limit is $2$ .
c) False: $(\csc x)' = -\csc x\cot x$ .
d) False: the limit is $0$ ; it is $\frac{1 - \cos x}{x^2}$ that tends to $\frac{1}{2}$ .
e) False: the quotient rule gives $(\tan x)' = \sec^2 x$ .

a)

FALSE. The proof of $(\sin x)' = \cos x$ uses $\frac{\sin h}{h} \to 1$ , true in radians only. With $S(x) = \sin(x^\circ)$ , the addition formula gives $\frac{S(x + h) - S(x)}{h} = \sin(x^\circ)\cdot\frac{\cos(h^\circ) - 1}{h} + \cos(x^\circ)\cdot\frac{\sin(h^\circ)}{h}$ . By Exercise 1 d), $\frac{\sin(h^\circ)}{h} \to \frac{\pi}{180}$ , and the same substitution $u = \frac{\pi h}{180}$ gives $\frac{\cos(h^\circ) - 1}{h} \to 0$ . So $S'(x) = \frac{\pi}{180}\cos(x^\circ)$ . At $x = 0$ the slope is $\frac{\pi}{180}$ , not $1$ : the graph of $\sin(x^\circ)$ is the graph of $\sin x$ stretched horizontally by the factor $\frac{180}{\pi}$ , period $360$ instead of $2\pi$ , and correspondingly flatter. Correct statement: $(\sin x)' = \cos x$ with $x$ in RADIANS.

b)

FALSE. The standard limit needs the same box in the numerator and, squared, in the denominator. Here the box is $2x$ , so $\frac{1 - \cos 2x}{x^2} = 4\cdot\frac{1 - \cos 2x}{(2x)^2} \to 4 \cdot \frac{1}{2} = 2$ . Second road, by the double angle identity $1 - \cos 2x = 2\sin^2 x$ : $\frac{2\sin^2 x}{x^2} = 2\left(\frac{\sin x}{x}\right)^2 \to 2$ . Plausibility: near $0$ , $1 - \cos u$ is about $\frac{u^2}{2}$ , so $1 - \cos 2x$ is about $2x^2$ and the quotient about $2$ . Correct statement: $\lim_{x\to 0}\frac{1 - \cos ax}{x^2} = \frac{a^2}{2}$ for every constant $a$ .

c)

FALSE. $(\csc x)' = \left(\frac{1}{\sin x}\right)' = \frac{-\cos x}{\sin^2 x} = -\csc x\cot x$ . A sign test settles it without any formula: on $\left(0, \frac{\pi}{2}\right)$ , $\csc x = \frac{1}{\sin x}$ decreases from very large values down to $1$ as $\sin x$ grows to $1$ , so its slope must be negative, while $\csc x\cot x > 0$ there. At $\frac{\pi}{4}$ the true slope is $-\sqrt 2\cdot 1 = -\sqrt 2$ . Correct statement: $(\csc x)' = -\csc x\cot x$ ; every co-function takes a minus sign.

d)

FALSE. Multiplying by the conjugate, $\frac{1 - \cos x}{x} = \frac{\sin x}{x}\cdot\frac{\sin x}{1 + \cos x} \to 1 \cdot 0 = 0$ . The numerator $1 - \cos x$ shrinks like $\frac{x^2}{2}$ , much faster than $x$ , while $\sin x$ shrinks like $x$ itself. Correct statement: $\lim_{x\to 0}\frac{1 - \cos x}{x} = 0$ and $\lim_{x\to 0}\frac{1 - \cos x}{x^2} = \frac{1}{2}$ . The limit $0$ is exactly what makes the term $\sin x\cdot\frac{\cos h - 1}{h}$ disappear in the proof of $(\sin x)' = \cos x$ : a $1$ there would give the wrong derivative.

e)

FALSE. There is no quotient-of-derivatives rule, any more than a product-of-derivatives rule. The quotient rule gives $(\tan x)' = \frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$ . Counterexamples to the student's answer: at $x = 0$ , $-\cot x$ is not even defined, while the tangent to $y = \tan x$ at the origin is $y = x$ , slope $1$ ; at $\frac{\pi}{4}$ the student gets $-1$ , the truth is $2$ . Correct statement: $(\tan x)' = \sec^2 x = 1 + \tan^2 x$ .

Exercise 9: A mass on a spring: velocity, acceleration and the true amplitude

A mass hangs at the end of a spring. Its position, measured in centimetres from the equilibrium position and positive in the direction that stretches the spring, is $x(t) = 3\cos t + 4\sin t$ , with $t$ in seconds and $t \ge 0$ . The mass was pulled to $x = 3$ and, instead of being released from rest, was given a push at $t = 0$ .

The figure shows $x(t)$ for the first seven seconds, the dot marking the start. No calculator: every time and every position must be exact.

a) Find $x(0)$ , the velocity $v(t) = x'(t)$ and $v(0)$ . Describe in words how the motion starts.
b) Find the acceleration $a(t) = v'(t)$ and show that $a(t) = -x(t)$ for every $t$ . What does this say about the direction of the acceleration?
c) Find the first time $t_1 > 0$ at which the mass is momentarily at rest, and its position then. A student reads the amplitude as $3$ cm, the starting position: correct him.
d) Prove that $x(t)^2 + v(t)^2 = 25$ for every $t$ . Deduce the largest distance from equilibrium and the largest speed, and say where the mass is when each occurs.
e) At $t = \frac{\pi}{2}$ , is the mass moving toward or away from equilibrium? Is it speeding up or slowing down?

Show the solution

Answers

a) $x(0) = 3$ cm; $v(t) = -3\sin t + 4\cos t$ ; $v(0) = 4$ cm/s, moving away from equilibrium.
b) $a(t) = -3\cos t - 4\sin t = -x(t)$ : always directed toward equilibrium.
c) $t_1 = \arctan\frac{4}{3}$ s, where $x(t_1) = 5$ cm: the amplitude is $5$ cm.
d) $|x| \le 5$ cm, reached where $v = 0$ ; $|v| \le 5$ cm/s, reached at equilibrium $x = 0$ .
e) $x = 4$ , $v = -3$ , $a = -4$ : toward equilibrium, speeding up.

a)

$x(0) = 3\cos 0 + 4\sin 0 = 3$ cm. By the sum and constant multiple rules, $v(t) = x'(t) = -3\sin t + 4\cos t$ , in centimetres per second, and $v(0) = 0 + 4 = 4$ cm/s. At the start the mass is $3$ cm on the stretched side and moves at $4$ cm/s AWAY from equilibrium, in the positive direction: the push was outward, and the mass will go beyond $3$ cm before turning back. The figure shows the curve rising from the dot.

b)

$a(t) = v'(t) = -3\cos t - 4\sin t = -(3\cos t + 4\sin t) = -x(t)$ . The acceleration always has the sign opposite to the position: on the stretched side ( $x > 0$ ) it points back toward equilibrium, on the compressed side it points back as well, and it is proportional to the distance, zero exactly at equilibrium. This is the signature of the motion of a mass on a spring, verified here from the formula by two differentiations; nothing is solved, the relation is checked.

c)

At rest means $v(t) = 0$ : $4\cos t = 3\sin t$ . If $\cos t = 0$ then $\sin t = \pm 1$ and the equation fails, so $\cos t \ne 0$ and dividing is safe: $\tan t = \frac{4}{3}$ . The solutions are $t = \arctan\frac{4}{3} + k\pi$ , and the first positive one is $t_1 = \arctan\frac{4}{3}$ , in $\left(0, \frac{\pi}{2}\right)$ . In that quadrant, a right triangle with opposite side $4$ and adjacent side $3$ has hypotenuse $5$ , so $\sin t_1 = \frac{4}{5}$ and $\cos t_1 = \frac{3}{5}$ . Then $x(t_1) = 3\cdot\frac{3}{5} + 4\cdot\frac{4}{5} = \frac{9 + 16}{5} = 5$ cm. The student's $3$ cm is where the mass STARTED, not where it turns: the push carried it $2$ cm further. The amplitude is found where the velocity vanishes, which is what the derivative is for.

d)

Expand: $x^2 = 9\cos^2 t + 24\sin t\cos t + 16\sin^2 t$ and $v^2 = 16\cos^2 t - 24\sin t\cos t + 9\sin^2 t$ . The cross terms cancel and $x^2 + v^2 = 25(\cos^2 t + \sin^2 t) = 25$ . Consequences: $x^2 = 25 - v^2 \le 25$ , so $|x| \le 5$ , with equality exactly when $v = 0$ , at the turning points; and $v^2 = 25 - x^2 \le 25$ , so $|v| \le 5$ , with equality exactly when $x = 0$ , as the mass passes through equilibrium. Largest distance $5$ cm, largest speed $5$ cm/s. The identity is the conservation of energy in disguise: position and speed trade off along a circle of radius $5$ .

e)

$x\left(\frac{\pi}{2}\right) = 3 \cdot 0 + 4 \cdot 1 = 4$ , $v\left(\frac{\pi}{2}\right) = -3 \cdot 1 + 4 \cdot 0 = -3$ , $a\left(\frac{\pi}{2}\right) = -4$ . The mass is on the positive side and its velocity is negative: it moves TOWARD equilibrium. Velocity and acceleration are both negative, so they have the same sign: the speed $|v|$ is increasing, the mass speeds up. Both answers agree with d): it is heading for $x = 0$ , where its speed will reach its largest value $5$ cm/s. Check: $4^2 + (-3)^2 = 25$ . Speeding up is decided by the signs of $v$ and $a$ TOGETHER, never by the sign of $a$ alone.

Exercise 10: A final exam question: the function sin x over (2 + cos x)

Let $f(x) = \frac{\sin x}{2 + \cos x}$ , whose graph on $[0, 2\pi]$ is shown in the figure. This is the shape of a long final exam question: domain and symmetries, a limit, a derivative simplified by an identity, horizontal tangents, then tangent lines, every step justified and every value exact.

a) Show that $f$ is defined on all of $\mathbb{R}$ , that it is $2\pi$ -periodic and that it is odd.
b) Find $\lim_{x\to 0}\frac{f(x)}{x}$ in two ways: with the limit of $\frac{\sin x}{x}$ , then by recognizing a derivative.
c) Compute $f'(x)$ and simplify it to $\frac{2\cos x + 1}{(2 + \cos x)^2}$ .
d) Find the points of the graph on $[0, 2\pi]$ where the tangent is horizontal, with exact coordinates, and check them against the symmetries of a).
e) Find the tangent lines at $x = 0$ and at $x = \pi$ , and the point where they meet.

Show the solution

Answers

a) $2 + \cos x \ge 1 > 0$ ; $f(x + 2\pi) = f(x)$ ; $f(-x) = -f(x)$
b) $\frac{1}{3}$ , which is $f'(0)$
c) $f'(x) = \frac{2\cos x + 1}{(2 + \cos x)^2}$
d) $\left(\frac{2\pi}{3}, \frac{\sqrt 3}{3}\right)$ and $\left(\frac{4\pi}{3}, -\frac{\sqrt 3}{3}\right)$
e) $y = \frac{x}{3}$ and $y = \pi - x$ , meeting at $\left(\frac{3\pi}{4}, \frac{\pi}{4}\right)$

a)

Since $\cos x \ge -1$ , the denominator satisfies $2 + \cos x \ge 1 > 0$ : it never vanishes, and $f$ , a quotient of continuous functions, is defined for every real $x$ . Sine and cosine have period $2\pi$ , so $f(x + 2\pi) = \frac{\sin x}{2 + \cos x} = f(x)$ . And $f(-x) = \frac{\sin(-x)}{2 + \cos(-x)} = \frac{-\sin x}{2 + \cos x} = -f(x)$ , because sine is odd and cosine is even: $f$ is odd, its graph symmetric about the origin.

b)

First way: $\frac{f(x)}{x} = \frac{\sin x}{x}\cdot\frac{1}{2 + \cos x} \to 1 \cdot \frac{1}{2 + 1} = \frac{1}{3}$ , by the product law and the continuity of cosine. Second way: $f(0) = 0$ , so $\frac{f(x)}{x} = \frac{f(x) - f(0)}{x - 0}$ , the difference quotient of $f$ at $0$ , whose limit is by definition $f'(0)$ . With c), $f'(0) = \frac{2 + 1}{(2 + 1)^2} = \frac{1}{3}$ . The two answers agree, which is a check on c) before it is even written.

c)

Quotient rule with $u = \sin x$ , $v = 2 + \cos x$ , $u' = \cos x$ , $v' = -\sin x$ : $f'(x) = \frac{\cos x(2 + \cos x) - \sin x(-\sin x)}{(2 + \cos x)^2} = \frac{2\cos x + \cos^2 x + \sin^2 x}{(2 + \cos x)^2} = \frac{2\cos x + 1}{(2 + \cos x)^2}$ , by $\cos^2 x + \sin^2 x = 1$ . The mark for this question is in the middle line: the two minus signs made a plus, and the identity was applied.

d)

The denominator of $f'$ is at least $1$ , so $f'(x) = 0 \iff 2\cos x + 1 = 0 \iff \cos x = -\frac{1}{2}$ , that is $x = \frac{2\pi}{3}$ or $x = \frac{4\pi}{3}$ on $[0, 2\pi]$ . Heights: $f\left(\frac{2\pi}{3}\right) = \frac{\sqrt 3/2}{2 - 1/2} = \frac{\sqrt 3/2}{3/2} = \frac{\sqrt 3}{3}$ , and $f\left(\frac{4\pi}{3}\right) = \frac{-\sqrt 3/2}{3/2} = -\frac{\sqrt 3}{3}$ . Check with a): $\frac{4\pi}{3} = 2\pi - \frac{2\pi}{3}$ , so by periodicity and oddness $f\left(\frac{4\pi}{3}\right) = f\left(-\frac{2\pi}{3}\right) = -f\left(\frac{2\pi}{3}\right)$ , as found. The figure agrees: a crest a little after $\frac{\pi}{2}$ and a trough a little before $\frac{3\pi}{2}$ , at the same distance from the axis.

e)

At $0$ : $f(0) = 0$ and $f'(0) = \frac{1}{3}$ , so the tangent is $y = \frac{x}{3}$ . At $\pi$ : $f(\pi) = \frac{0}{2 - 1} = 0$ and $f'(\pi) = \frac{2(-1) + 1}{(2 - 1)^2} = -1$ , so the tangent is $y = -(x - \pi) = \pi - x$ . They meet where $\frac{x}{3} = \pi - x$ , that is $\frac{4x}{3} = \pi$ , so $x = \frac{3\pi}{4}$ and $y = \frac{\pi}{4}$ . The meeting point $\left(\frac{3\pi}{4}, \frac{\pi}{4}\right)$ lies ABOVE the graph: the height of the graph at $\frac{3\pi}{4}$ is $\frac{\sqrt 2/2}{2 - \sqrt 2/2} = \frac{\sqrt 2}{4 - \sqrt 2}$ , which is less than $\frac{3}{4}$ because $4\sqrt 2 < 12 - 3\sqrt 2$ , that is $7\sqrt 2 < 12$ , or $98 < 144$ after squaring, while $\frac{\pi}{4} > \frac{3}{4}$ . The solution figure shows the two tangents and their meeting point.

Corrected exercises: derivatives of trigonometric functions (MATH 140)

Course recap

Part A: the basics (/50)

Exercise 1: Where the derivative of sine comes from: the limit of sin t over t

Exercise 2: Limits built on sin(u)/u and 1 - cos(u): match the box before you cancel

Exercise 3: The six derivatives: two proved from the definition, four by the quotient rule

Exercise 4: Product and quotient rules with trigonometric functions, then simplify

Exercise 5: Derivatives of high order: the cycle of four

Part B: problems and reasoning (/50)

Exercise 6: Horizontal tangents on an interval: factor, never divide

Exercise 7: Tangent and normal lines to trigonometric curves

Exercise 8: Five statements to correct

Exercise 9: A mass on a spring: velocity, acceleration and the true amplitude

Exercise 10: A final exam question: the function sin x over (2 + cos x)

See also

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