Problems
Problem 1
On \(0\le x\le2\pi\), let \(A(x)=\arccos(\cos x)\).
A student cancels the two functions and concludes \(A(x)=x\) and \(A^{\prime}(x)=1\) throughout the interval. Find exactly where that conclusion fails. Give a piecewise formula for \(A\) and use its graph to explain how composing two familiar continuous functions can create a corner.
Hint
Arccosine always returns an angle in \([0,\pi]\). For the second half of the interval, use \(\cos(2\pi-x)=\cos x\).
Solution
The principal branch forces
\[A(x)=\begin{cases}x,&0\le x\le\pi,\\2\pi-x,&\pi\le x\le2\pi.\end{cases}\]Its graph consists of the segments from \((0,0)\) to \((\pi,\pi)\) and then to \((2\pi,0)\).
Thus
\[A^{\prime}(x)=\begin{cases}1,&0<x<\pi,\\-1,&\pi<x<2\pi.\end{cases}\]At \(\pi\) the one-sided slopes differ, so no derivative exists. Directly substituting into the chain-rule expression \(\sin x/\sqrt{1-\cos^2x}\) there gives an undefined quotient, not a justification for assigning a slope.
Problem 2
Can a function grow without bound while its slope always stays between \(-1\) and \(1\)? Decide using
\[L(x)=\ln(e^x+e^{-x}).\]Find its derivative, give a bound for it, and examine what happens to \(L(x)\) as \(x\to+\infty\).
Hint
Compare the sum and difference of the two positive numbers \(e^x\) and \(e^{-x}\). For the limit, factor \(e^x\) inside the logarithm.
Solution
The logarithm rule gives
\[L^{\prime}(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\tanh x.\]For positive \(u,v\), \(\lvert u-v\rvert<u+v\). Therefore
\[-1<L^{\prime}(x)<1.\]But
\[L(x)=\ln\bigl(e^x(1+e^{-2x})\bigr)=x+\ln(1+e^{-2x})\longrightarrow+\infty.\]So the answer is yes. A bound on how fast a function changes is not a bound on how large it can become.
Problem 3
A sensor reading, in concentration units, is modelled by
\[C(t)=5+40e^{-kt},\qquad k>0,\]with time in hours. A measurement gives \(C(2)=25\).
The reading falls from \(45\) to \(25\) in two hours. A student says, “The amount has not halved, so this cannot be exponential decay.” What has the student overlooked?
Find \(k\) and the rate at which the reading is falling after two hours. State exactly which quantity halves every two hours.
Hint
First isolate the exponential in the measurement equation. The constant background contributes to the reading but not to its derivative.
Solution
The measurement gives \(20=40e^{-2k}\), hence
\[k=\frac{\ln2}{2}\,\mathrm{h}^{-1}.\]Differentiation yields \(C^{\prime}(t)=-40ke^{-kt}\), and therefore
\[C^{\prime}(2)=-20k=-10\ln2\]concentration units per hour. The constant \(5\) is a background reading. The quantity above background is
\[E(t)=C(t)-5=40e^{-kt}.\]Since \(e^{-2k}=1/2\),
\[E(t+2)=E(t)e^{-2k}=\frac12E(t).\]It is the amount above background that halves every two hours, not the full reading. A fixed background can hide the halving pattern in the measured numbers.
Problem 4
For \(P(x)=(2+\sin x)^{\cos x}\), a student says: “Whenever the exponent is zero, the function is \(1\), so its derivative must be zero too.” Decide whether this reasoning is valid at \(x=\pi/2\). Compare with \(x=0\) by finding both tangent lines. At each input, identify which changing part—the base or the exponent—contributes to the slope.
Hint
The base is always positive. Differentiate \(\ln P=\cos(x)\ln(2+\sin x)\) using the product rule.
Solution
Logarithmic differentiation gives
\[\frac{P^{\prime}(x)}{P(x)}=-\sin(x)\ln(2+\sin x)+\frac{\cos^2x}{2+\sin x}.\]At zero, \(P(0)=2\) and \(P^{\prime}(0)=2(1/2)=1\). The tangent is
\[y=2+x.\]At \(\pi/2\), \(P(\pi/2)=1\) and \(P^{\prime}(\pi/2)=-\ln3\). Thus
\[y=1-\ln3\left(x-\frac\pi2\right).\]At zero, the exponent has zero derivative, and only the changing base contributes. At \(\pi/2\), the base has zero derivative, and only the changing exponent contributes.
Holding the exponent fixed at zero would turn the expression into the constant \(1\) and incorrectly predict zero slope. Here the entire nonzero rate comes from the changing exponent.
Problem 5
A lamp is \(4\,\mathrm m\) from a straight wall. Its beam makes angle \(\theta\) with the perpendicular to the wall. The light spot is at position \(s=4\tan\theta\) along the wall.
\(\theta\)
\(4\,\mathrm m\)
\(s\)
The beam turns at a constant rate of \(0.2\,\mathrm{rad/s}\), starting at \(\theta=0\). Will the spot cover equal distances in equal times? Find when its speed first doubles, and find a turning rate \(d\theta/dt\) that would instead make the spot move at the constant speed \(1\,\mathrm{m/s}\).
Hint
The chain rule connects \(ds/dt\) and \(d\theta/dt\). For the final part, work backwards from the desired value of \(ds/dt\).
Solution
The chain rule gives
\[\frac{ds}{dt}=4\sec^2\theta\frac{d\theta}{dt}.\]For the constant turning rate, this is
\[\frac{ds}{dt}=\frac45\sec^2\theta.\]The spot does not move at constant speed. Its initial speed is \(4/5\,\mathrm{m/s}\), and this first doubles when \(\sec^2\theta=2\), at \(\theta=\pi/4\). Since \(\theta=t/5\), the time is \(5\pi/4\) seconds.
To get \(ds/dt=1\,\mathrm{m/s}\) instead, we must choose
\[\frac{d\theta}{dt}=\frac14\cos^2\theta\quad\mathrm{rad/s}.\]The beam must turn more slowly as it points farther along the wall. Working backwards through the chain rule tells us how to produce a desired motion.
Problem 6
A camera is \(3\,\mathrm m\) from a straight track. An object occupies the segment from coordinate \(s\) to \(s+2\) along the track, measured in metres from the point nearest the camera. Its angular width is
\[\alpha(s)=\arctan\!\left(\frac{s+2}{3}\right)-\arctan\!\left(\frac{s}{3}\right).\]Two positions symmetric about \(s=-1\) have the same angular width. Must they have the same rate of change of angular width? Show why the widths agree, then compare their derivatives. Where is the derivative zero?
Hint
Simplify each derivative to a fraction with numerator \(3\). Compare the two positive denominators.
Solution
Arctangent is odd, so
\[\alpha(-2-s)=\arctan(-s/3)-\arctan(-(s+2)/3)=\alpha(s).\]The inverse tangent rule and chain rule give
\[\alpha^{\prime}(s)=\frac3{9+(s+2)^2}-\frac3{9+s^2} =\frac{-12(s+1)}{(9+(s+2)^2)(9+s^2)}.\]The denominator is positive. The rate is positive for \(s<-1\), zero at \(s=-1\), and negative for \(s>-1\), in radians per metre.
At \(s=-1\) the object occupies \([-1,1]\) and is centred opposite the camera. The angular width has a horizontal tangent as a function of position there. Differentiating the symmetry identity also gives
\[-\alpha^{\prime}(-2-s)=\alpha^{\prime}(s).\]The symmetric positions have equal widths but opposite rates. Equal function values do not imply equal derivatives.
Problem 7
Two students differentiate
\[F(x)=\ln\!\left(\frac{1-\cos x}{\sin x}\right),\qquad 0<x<\pi.\]One first writes the fraction as \(\csc x-\cot x\). The other first separates the logarithm into a difference of logarithms. Use both routes and show that the apparently different answers agree.
What must you check before separating the logarithm?
Hint
On the given interval, both numerator and denominator are positive. In the second route, use \(\sin^2x+\cos^2x=1\) to combine the result.
Solution
Since \(1-\cos x>0\) and \(\sin x>0\) on \(0<x<\pi\), both routes are valid.
The first route gives
\[F^{\prime}(x)=\frac{-\csc x\cot x+\csc^2x}{\csc x-\cot x}=\csc x.\]The second route starts from
\[F(x)=\ln(1-\cos x)-\ln(\sin x),\]and gives
\[\begin{aligned} F^{\prime}(x)&=\frac{\sin x}{1-\cos x}-\frac{\cos x}{\sin x}\\ &=\frac{\sin^2x-\cos x(1-\cos x)}{\sin x(1-\cos x)}\\ &=\frac{1-\cos x}{\sin x(1-\cos x)}=\csc x. \end{aligned}\]The answers agree. The useful choice is often how to rewrite the function before differentiating; different-looking derivative formulas need not disagree. Positivity justifies taking each separate logarithm.
Problem 8
A differentiable branch of the curve
\[e^{xy}+x+y=e+2\]passes through \((1,1)\). Find its tangent line and its second derivative there. Why does the exponential term have zero first derivative at this point even though neither coordinate is zero?
Hint
Keep \(y+xy^{\prime}\) as a single factor after the first differentiation. It represents the rate of change of the exponent.
Solution
The point satisfies the equation. Differentiating gives
\[e^{xy}(y+xy^{\prime})+1+y^{\prime}=0.\]At \((1,1)\) this becomes \((e+1)(1+y^{\prime})=0\), so \(y^{\prime}=-1\) and the tangent is
\[y-1=-(x-1).\]The exponent’s derivative is \(y+xy^{\prime}=0\) there: the two contributions cancel. Differentiate again before substituting:
\[e^{xy}\left[(y+xy^{\prime})^2+2y^{\prime}+xy^{\prime\prime}\right]+y^{\prime\prime}=0.\]At the point,
\[e(-2+y^{\prime\prime})+y^{\prime\prime}=0,\qquad y^{\prime\prime}=\frac{2e}{e+1}.\]A term can have zero first derivative at a point while having a nonzero second derivative.
Problem 9
A strictly increasing, twice differentiable function satisfies
\[f(2)=5,\qquad f^{\prime}(2)=3,\qquad f^{\prime\prime}(2)=-6.\]Let \(g=f^{-1}\) near \(5\). A student takes reciprocals twice and predicts \(g^{\prime\prime}(5)=1/f^{\prime\prime}(2)\). Find the error by deriving a formula for the second derivative of an inverse. Calculate the two derivatives of \(g\) at \(5\), and decide whether the original slope and inverse slope are changing in the same direction at the corresponding inputs.
Hint
Start with \(f(g(y))=y\) and differentiate twice. Keep the factor from the inner derivative on both differentiations.
Solution
The identity gives
\[f^{\prime}(g(y))g^{\prime}(y)=1,\]so \(g^{\prime}(5)=1/3\). Differentiating once more,
\[f^{\prime\prime}(g(y))(g^{\prime}(y))^2+f^{\prime}(g(y))g^{\prime\prime}(y)=0.\]Consequently,
\[g^{\prime\prime}(5)=-\frac{f^{\prime\prime}(2)}{(f^{\prime}(2))^3}=\frac{6}{27}=\frac29.\]In general, the calculation shows
\[g^{\prime\prime}(y)=-\frac{f^{\prime\prime}(g(y))}{(f^{\prime}(g(y)))^3}.\]The second derivative is not the reciprocal of the original second derivative. Differentiating the first reciprocal rule requires both a minus sign and the inner derivative.
Here the original slope is decreasing at \(2\) because \(f^{\prime\prime}(2)<0\), while the inverse slope is increasing at \(5\) because \(g^{\prime\prime}(5)>0\). Reflection swaps the roles of the variables, rather than preserving how the slope changes.
Problem 10
For a real input \(u\), define
\[H(u)=\ln\!\left(u+\sqrt{1+u^2}\right).\]- Explain why its logarithm is defined for every real \(u\).
- Show directly from the exponential definitions that \(\sinh(H(u))=u\).
- Find \(H^{\prime}(u)\) both directly and by the inverse derivative rule.
Hint
The two factors \(\sqrt{1+u^2}+u\) and \(\sqrt{1+u^2}-u\) have product \(1\).
Solution
Since \(\sqrt{1+u^2}>\lvert u\rvert\), the logarithm’s argument is positive. Put \(v=H(u)\). Then
\[e^v=u+\sqrt{1+u^2},\qquad e^{-v}=\sqrt{1+u^2}-u.\]Thus
\[\sinh v=\frac{e^v-e^{-v}}2=u,\qquad \cosh v=\sqrt{1+u^2}.\]Direct differentiation gives
\[H^{\prime}(u)=\frac{1+u/\sqrt{1+u^2}}{u+\sqrt{1+u^2}}=\frac1{\sqrt{1+u^2}}.\]Alternatively, differentiate \(\sinh(H(u))=u\) to obtain \(\cosh(H(u))H^{\prime}(u)=1\) and the same answer. Since \(\sinh\) is continuous, strictly increasing, and has range \(\mathbb R\), \(H\) is its inverse. This problem derives that inverse rather than assuming a new derivative formula.
Problem 11
Let
\[R(x)=\frac{x^2e^{-x}}{1+x},\qquad x>0.\]Find the positive input \(a\) where \(R^{\prime}(a)=0\). Then determine whether \(R^{\prime\prime}(a)\) is zero, positive, or negative, without expanding a large quotient derivative. Does the horizontal tangent mean that \(R\) is constant near \(a\)?
Hint
Use logarithmic differentiation. If \(q=R^{\prime}/R\), differentiate \(R^{\prime}=Rq\). At the required input, \(q=0\).
Solution
All factors are positive on the domain. Taking logarithms gives
\[\ln R=2\ln x-x-\ln(1+x),\]so
\[q(x)=\frac{R^{\prime}(x)}{R(x)}=\frac2x-1-\frac1{1+x}=\frac{2-x^2}{x(1+x)}.\]Because \(R(x)>0\), its derivative is zero exactly when \(q(x)=0\). Thus \(a=\sqrt2\).
Differentiate \(R^{\prime}=Rq\):
\[R^{\prime\prime}=R^{\prime}q+Rq^{\prime}=R(q^2+q^{\prime}).\]At \(a\), \(q(a)=0\), so
\[\frac{R^{\prime\prime}(a)}{R(a)}=q^{\prime}(a) =-1+\frac1{(1+\sqrt2)^2}=2-2\sqrt2<0.\]Since \(R(a)>0\), \(R^{\prime\prime}(a)<0\). The function is not constant near \(a\): if it were, its second derivative there would also be zero. Taking logarithms can simplify a second-derivative question as well as a first-derivative calculation.
Problem 12
Consider the part of the implicit curve
\[x^3+y^3=6xy\]with \(x>0\) and \(y>0\).
- Find every horizontal and vertical tangent on this part of the curve, with exact coordinates and tangent equations.
- Find the tangent and \(y^{\prime\prime}\) at \((3,3)\).
Hint
Differentiate implicitly, and solve the curve equation together with the numerator-zero condition or denominator-zero condition. A vertical tangent can be checked by finding \(dx/dy\).
Solution
Differentiation gives
\[(x^2-2y)+(y^2-2x)y^{\prime}=0,\qquad y^{\prime}=\frac{2y-x^2}{y^2-2x}.\]For a horizontal tangent, \(y=x^2/2\). Substituting into the curve yields
\[x^3+\frac{x^6}{8}=3x^3,\qquad x^3=16\]because \(x>0\). Put \(r=\sqrt[3]{2}\). The point is \((2r,2r^2)\), where \(y^2-2x=4r\ne0\). Its tangent is \(y=2r^2\).
By symmetry, the vertical tangent is at \((2r^2,2r)\), with line \(x=2r^2\). There \(dx/dy=(2x-y^2)/(x^2-2y)=0\), with nonzero denominator, confirming the vertical direction. These equations have no other positive solutions.
At \((3,3)\) the slope is \(-1\), so the tangent is \(x+y=6\). Differentiate the unsolved first-derivative equation:
\[2x-2y^{\prime}+(2yy^{\prime}-2)y^{\prime}+(y^2-2x)y^{\prime\prime}=0.\]Substitution gives \(8+8+3y^{\prime\prime}=0\), hence
\[y^{\prime\prime}(3)=-\frac{16}{3}.\]
At the origin, both numerator and denominator of the slope formula vanish. This calculation alone says nothing about a tangent there; the positive-coordinate restriction keeps that separate question out of this problem.
Problem 13
For positive constants \(p,q\), define
\[f(x)=\begin{cases}\lvert x\rvert^p\sin(\lvert x\rvert^{-q}),&x\ne0,\\0,&x=0.\end{cases}\]Find exactly which pairs \(p,q\) make each statement true:
- \(f\) is differentiable at zero.
- \(f\) is differentiable at zero and \(f^{\prime}\) is continuous there.
- \(f^{\prime}(0)=0\), but the slopes at nearby points are unbounded.
Prove both directions of each claim. Give one concrete function for each of the three possible behaviours: a continuous derivative, a bounded but discontinuous derivative, and an unbounded derivative near zero. No sketch is required.
Hint
Separate the small size of the wave from its rapid oscillation. First study the difference quotient at zero. For nearby slopes, differentiate on \(x>0\) and choose inputs where the sine is zero or where it is one. A bound proves sufficiency; carefully chosen inputs prove necessity.
Solution
The function is continuous at zero for every \(p>0\), because \(\lvert f(x)\rvert\le\lvert x\rvert^p\). For the derivative, the difference quotient satisfies
\[\left\lvert\frac{f(h)}h\right\rvert\le\lvert h\rvert^{p-1}.\]Thus \(p>1\) gives \(f^{\prime}(0)=0\). This condition is necessary. Along positive inputs
\[h_n=(\pi/2+2n\pi)^{-1/q},\]the quotient is \(h_n^{p-1}\). For \(p<1\) it grows without bound; for \(p=1\) it equals \(1\), while at \(h_n=(2n\pi)^{-1/q}\) it is zero. Therefore
\[f\text{ is differentiable at }0\quad\Longleftrightarrow\quad p>1.\]For \(x>0\),
\[f^{\prime}(x)=px^{p-1}\sin(x^{-q})-qx^{p-q-1}\cos(x^{-q}).\]For \(x<0\), evenness of \(f\) makes its derivative the negative of the derivative at \(-x\). It is enough to bound the positive side.
If \(p>q+1\), both terms tend to zero, so the derivative is continuous at zero. Conversely, at \(x_n=(2n\pi)^{-1/q}\),
\[f^{\prime}(x_n)=-qx_n^{p-q-1}.\]When \(p=q+1\) this equals \(-q\), and when \(p<q+1\) its magnitude grows without bound. For a differentiable function in this family,
\[f^{\prime}\text{ is continuous at }0\quad\Longleftrightarrow\quad p>q+1.\]When \(p=q+1\), the displayed derivative formula is bounded near zero, yet is not continuous there. When \(1<p<q+1\), the same special inputs show unbounded slopes. Hence the answer to the third question is exactly
\[1<p<q+1.\]Examples, each with derivative zero at the origin, are
- \(\lvert x\rvert^3\sin(1/\lvert x\rvert)\): continuous derivative;
- \(x^2\sin(1/\lvert x\rvert)\): bounded but discontinuous derivative;
- \(x^2\sin(1/x^2)\): derivative unbounded near zero.
Each is assigned value zero at zero. A wave can shrink enough to have a tangent at the origin while oscillating too quickly for nearby tangents to settle down.
Problem 14
How many straight lines are tangent to both \(y=e^x\) and \(y=\ln x\)?
Find all such lines, giving their points of tangency, and prove that none are missing. An exact answer may define a constant as the unique positive solution of an equation; a decimal search alone is not a proof.
The two tangency points need not have the same input or be reflections of one another.
Hint
Call the common slope \(m\). Express each point of tangency and each tangent’s intercept in terms of \(m\). Equating the intercepts leaves one equation. To count its solutions, try \(t=\ln m\) and use the exponential definition of hyperbolic tangent.
Solution
Both derivatives are positive, so a common tangent must have slope \(m>0\).
For the exponential, \(e^a=m\) gives the point \((\ln m,m)\) and tangent
\[y=mx+m-m\ln m.\]For the logarithm, \(1/b=m\) gives the point \((1/m,-\ln m)\) and tangent
\[y=mx-1-\ln m.\]These are the same line exactly when
\[(m-1)\ln m=m+1.\]The value \(m=1\) does not satisfy the equation. Put \(t=\ln m\). Rearranging gives
\[t\frac{e^t-1}{e^t+1}=1,\qquad\text{or}\qquad t\tanh(t/2)=1.\]The expression on the left is even. On \(t>0\), both factors are positive and strictly increasing: the second is
\[\tanh(t/2)=1-\frac2{e^t+1},\]which increases because \(e^t\) increases. Their product is therefore strictly increasing, without needing a derivative test. It tends to zero as \(t\to0^+\) and grows without bound as \(t\to+\infty\). Continuity and the intermediate value theorem give exactly one positive solution, say \(t_0\). Evenness gives exactly one negative solution, \(-t_0\).
There are therefore exactly two common tangents. For each \(t\in\{t_0,-t_0\}\), the line and its two points are
\[y=e^t x-1-t,\qquad (t,e^t),\qquad(e^{-t},-t).\]Every common tangent was forced to have this form, and each solution produces a line tangent to both graphs. This proves completeness. The two slopes are reciprocals, as reflection across \(y=x\) suggests, but the two contact points on a single line are not a reflected pair.
