Problems
Problem 1
Let \(u(x)=x^3\) and \(v(u)=\sqrt[3]{u}\), using the real cube root for negative inputs too.
Find the derivative of \(v(u(x))\) at zero directly from the definition. Why is substituting \(x=0\) into \(v^{\prime}(u(x))u^{\prime}(x)\) not a valid way to obtain that answer?
Hint
Simplify the composition before differentiating. Check whether the outer function has a finite derivative at its input.
Solution
The composition is \(\sqrt[3]{x^3}=x\) for every real \(x\). Therefore,
\[\left.\frac{d}{dx}v(u(x))\right|_{x=0} =\lim_{h\to0}\frac{h-0}{h}=1.\]But the outer derivative at zero would require
\[v^{\prime}(0)=\lim_{k\to0}\frac{\sqrt[3]{k}}{k} =\lim_{k\to0}\frac1{|k|^{2/3}},\]which is not finite. The chain rule’s hypothesis fails at this input. An undefined factor multiplied by zero is not a derivative calculation—even though the composition itself is as simple as a straight line.
Problem 2
A sensor reports \(F(-1)=F(0)=F(1)=0\). Someone concludes that \(F^{\prime}(0)=0\).
Construct a polynomial satisfying all three measurements but with \(F^{\prime}(0)=5\). Then construct one with any prescribed slope \(m\) at zero. What do these examples say about recovering an exact instantaneous rate from finitely many values?
Hint
A multiple of \(x(x-1)(x+1)\) preserves all three measured zeros. Choose the multiplier to set the derivative.
Solution
Take \(F(x)=-5x(x^2-1)=5x-5x^3\). It vanishes at all three measured inputs, but
\[F^{\prime}(x)=5-15x^2,\qquad F^{\prime}(0)=5.\]For any real \(m\), use \(F_m(x)=mx-mx^3\). Then \(F_m^{\prime}(0)=m\) while the three sampled values remain zero.
These measurements alone do not fix the exact derivative, even among polynomials. Estimating a derivative from samples needs additional assumptions about behavior between measurements.
Problem 3
Suppose \(f\) is differentiable at \(2\), with \(f^{\prime}(2)=-3\). Find
\[\lim_{h\to0}\frac{f(2+3h)-f(2)}h, \qquad \lim_{h\to0}\frac{f(2+h)-f(2-h)}h.\]Explain why the second denominator is not the horizontal distance between the two sampled inputs.
Hint
For the first limit introduce the increment \(k=3h\). For the second, add and subtract \(f(2)\).
Solution
The first quotient is three times a derivative quotient with increment \(3h\):
\[\lim_{h\to0}3\frac{f(2+3h)-f(2)}{3h}=3f^{\prime}(2)=-9.\]For the second,
\[\begin{aligned} \lim_{h\to0}\frac{f(2+h)-f(2-h)}h &=\lim_{h\to0}\left[\frac{f(2+h)-f(2)}h+\frac{f(2)-f(2-h)}h\right]\\ &=f^{\prime}(2)+f^{\prime}(2)=-6. \end{aligned}\]The two inputs differ by \(2h\), not \(h\). Dividing by \(2h\) instead would give a limit of \(-3\).
Problem 4
The graph consists of the line segment from \((-2,-1)\) to \((0,1)\), followed by the parabola \(y=1+x^2\) for \(0\le x\le2\).
- Sketch \(f^{\prime}\) on \((-2,2)\), showing any excluded input with an open circle.
- Where does \(f^{\prime\prime}\) exist, and what are its values?
-
Changing only \(f(0)\) cannot repair the corner. Instead, find \(k\) such that $$g(x)=f(x)+k x \(is differentiable at zero. Does this also make\)g$$ twice differentiable there?
Hint
A height on the derivative graph represents a slope on this graph. Check the two limiting slopes at the join.
Solution
The line has slope \(1\), while the parabola has derivative \(2x\). Hence
\[f^{\prime}(x)=\begin{cases}1,&-2<x<0,\\2x,&0<x<2.\end{cases}\]Sketch a horizontal segment at height \(1\) to the left, with an open endpoint at \((0,1)\); to the right sketch \(y=2x\), with an open endpoint at \((0,0)\). No value is assigned at zero.
Differentiating these branches gives
\[f^{\prime\prime}(x)=\begin{cases}0,&-2<x<0,\\2,&0<x<2.\end{cases}\]The left and right slopes of \(f\) at zero are \(1\) and \(0\). Changing the value at the join cannot reconcile them. In fact, differentiability would require continuity, which already forces \(f(0)=1\).
For the proposed correction, the left slope is \(1-k\) and the right slope is \(k\). They match only when \(k=1/2\). Directly,
\[\lim_{h\to0^-}\frac{g(h)-g(0)}h=\frac12,\qquad \lim_{h\to0^+}\frac{g(h)-g(0)}h=\lim_{h\to0^+}(h+1/2)=\frac12.\]The repaired first derivative is \(1/2\) on the left and \(2x+1/2\) on the right, including value \(1/2\) at zero. Its left and right difference quotients at zero tend to \(0\) and \(2\), respectively. Thus the correction repairs the first derivative but not the second.
Problem 5
Find every tangent to \(y=x^3+2\) that passes through the origin. Give both the point of tangency and the line. Does it meet the curve at any other point?
Hint
Write the tangent at an unknown input \(a\), then impose the condition that \((0,0)\) lies on it.
Solution
At \(a\), the slope is \(3a^2\). The tangent is
\[y=a^3+2+3a^2(x-a)=3a^2x+2-2a^3.\]Passing through the origin requires \(2-2a^3=0\), so \(a=1\). The only such tangent is \(y=3x\), touching at \((1,3)\). To find other intersections, solve
\[x^3+2=3x\quad\Longleftrightarrow\quad (x-1)^2(x+2)=0.\]It also crosses the curve at \((-2,-6)\). At that input the curve’s slope is \(12\), not \(3\), so this second intersection is not another point of tangency.
A tangent can meet its curve elsewhere. Passing through a point and being tangent there are different conditions.
Problem 6
For \(Q(x)=(x^2-1)^2/(x-1)\), find \(Q^{\prime}(x)\) efficiently and state its domain. Can \(Q\) be extended to a differentiable function at \(1\)? If so, give the added value and the new tangent slope there.
Hint
Factor before differentiating. Distinguish the original function from its extension.
Solution
For \(x\ne1\),
\[Q(x)=(x-1)(x+1)^2,\]so the product rule gives
\[Q^{\prime}(x)=(x+1)^2+2(x-1)(x+1)=(x+1)(3x-1),\qquad x\ne1.\]The polynomial extension \(\widetilde Q(x)=(x-1)(x+1)^2\) is differentiable everywhere. Set \(\widetilde Q(1)=0\); its slope there is \(\widetilde Q^{\prime}(1)=4\). The original \(Q\) still has no value or derivative at \(1\).
Problem 7
A small light source moves away from a sensor. Its distance is \(r(t)=2+t^2\) metres, with \(t\ge0\) measured in seconds. The sensor model is \(I=120/r^2\), with \(I\) in sensor units.
- Find \(dI/dt\) at \(t=1\), including units.
- A student uses only \(dI/dr=-240/r^3\). What rate has the student actually calculated?
Hint
Distance is the intermediate variable. Keep the units of each derivative visible.
Solution
Since \(dr/dt=2t\),
\[\frac{dI}{dt}=\frac{dI}{dr}\frac{dr}{dt}=-\frac{240}{r^3}(2t)=-\frac{480t}{(2+t^2)^3}.\]At \(t=1\), this is \(-480/27=-160/9\) sensor units per second. The reading is decreasing.
The student has calculated the change in the reading per metre of distance, not per second. Multiplying by the distance rate converts between the two.
Problem 8
A robot moves along a straight line with position \(s(t)=t^3-3t^2\) metres for \(0\le t\le3\) seconds.
- When is it at rest, and when does it reverse direction within the time interval?
- Find its displacement and total distance travelled.
- At \(t=1\), is it stopped? What is its acceleration then?
Hint
Find the sign of velocity on either side of each zero. Total distance adds the lengths of the separate parts of the trip; it is not just the final position minus the initial position.
Solution
The velocity and acceleration are
\[v(t)=3t(t-2),\qquad a(t)=6t-6.\]It is at rest at \(t=0\) and \(t=2\). Velocity is negative for \(0<t<2\) and positive for \(2<t\le3\), so it reverses direction at \(2\). There is no earlier motion in this model from which to infer a reversal at the initial time.
The positions are \(s(0)=0\), \(s(2)=-4\), and \(s(3)=0\). Displacement is \(0\,\mathrm{m}\); distance is \(4+4=8\,\mathrm{m}\).
At \(t=1\), velocity is \(-3\,\mathrm{m/s}\) while acceleration is zero. Zero acceleration at an instant does not mean the robot is stopped.
Problem 9
A solution has dissolved mass \(M(t)\) grams and volume \(V(t)\) litres. Its concentration is \(C=M/V\). At one instant,
\[M=10,\quad M^{\prime}=-0.2,\quad V=2,\quad V^{\prime}=-0.1,\]with time in minutes. Is the concentration rising or falling, and how fast? Explain how your answer is compatible with dissolved mass being lost.
Hint
Differentiate the ratio. Losing some dissolved material and losing some liquid have competing effects on concentration.
Solution
The quotient rule gives
\[C^{\prime}=\frac{M^{\prime}V-MV^{\prime}}{V^2} =\frac{(-0.2)(2)-10(-0.1)}{2^2}=0.15\,\mathrm{g/(L\cdot min)}.\]The concentration is rising. The loss of volume concentrates what remains strongly enough to outweigh the loss of dissolved mass. The mass rate alone cannot determine the concentration rate.
Problem 10
Two twice differentiable functions satisfy
\[\begin{aligned} f(1)&=g(1)=1,\\ f^{\prime}(1)&=2,&g^{\prime}(1)&=3,\\ f^{\prime\prime}(1)&=-1,&g^{\prime\prime}(1)&=4. \end{aligned}\]Let \(A(x)=f(g(x))\) and \(B(x)=g(f(x))\). Show that they have the same value and tangent at \(1\), but different second derivatives there. Find both second derivatives and explain which step in the calculation prevents the two compositions from being interchangeable.
Hint
Differentiate a composition twice. The second differentiation acts on both factors of \(f^{\prime}(g(x))g^{\prime}(x)\).
Solution
Both values equal \(1\). The first derivatives at the common input are
\[A^{\prime}(1)=2\cdot3=6,\qquad B^{\prime}(1)=3\cdot2=6.\]Thus both tangents are \(y-1=6(x-1)\).
Applying the product and chain rules to the first derivative gives
\[A^{\prime\prime}(x)=f^{\prime\prime}(g(x))(g^{\prime}(x))^2 +f^{\prime}(g(x))g^{\prime\prime}(x).\]Consequently,
\[A^{\prime\prime}(1)=(-1)3^2+2\cdot4=-1.\]Swapping the functions gives
\[B^{\prime\prime}(1)=4\cdot2^2+3(-1)=13.\]The squared inner rate and the inner second derivative play different roles. Equal values and equal first derivatives at one input do not determine how the slopes change nearby.
Problem 11
Choose \(a,b,c\) so that
\[f(x)=\begin{cases} ax^2+bx+c,&x\le0,\\ \dfrac1{1+x},&x>0 \end{cases}\]has both a first and a second derivative at zero. For your choice, does the third derivative at zero exist? Use the limit definition at the join, rather than assigning values by substituting into formulas valid only away from zero.
Hint
Match the values, then the first difference quotients. Next do the same for the first derivative, and finally for the second derivative.
Solution
Continuity forces \(c=1\). On the two sides, the first difference quotients give
\[\lim_{h\to0^-}\frac{f(h)-1}{h}=\lim_{h\to0^-}(ah+b)=b,\] \[\lim_{h\to0^+}\frac{f(h)-1}{h} =\lim_{h\to0^+}\frac{-1}{1+h}=-1.\]Thus \(b=-1\) and \(f^{\prime}(0)=-1\). The left branch of \(f^{\prime}\) is \(2ax-1\), and the right branch is \(-(1+x)^{-2}\). Their difference quotients at zero satisfy
\[\lim_{h\to0^-}\frac{f^{\prime}(h)+1}{h}=2a,\] \[\lim_{h\to0^+}\frac{f^{\prime}(h)+1}{h} =\lim_{h\to0^+}\frac{2+h}{(1+h)^2}=2.\]Therefore \(a=1\) and \(f^{\prime\prime}(0)=2\). The second derivative equals \(2\) on the left and \(2/(1+x)^3\) on the right. Finally,
\[\lim_{h\to0^-}\frac{f^{\prime\prime}(h)-2}{h}=0,\] \[\begin{aligned} \lim_{h\to0^+}\frac{f^{\prime\prime}(h)-2}{h} &=\lim_{h\to0^+}\frac{-2(3+3h+h^2)}{(1+h)^3}=-6. \end{aligned}\]The third derivative does not exist. Matching two levels of derivatives does not automatically match the next.
Problem 12
Find every line tangent to the graph of \(f(x)=x^4-2x^2\) at two distinct points. Give the line and both points of tangency, and prove that your list is complete.
Hint
Let the two inputs be \(a\ne b\). The tangents need the same slope and the same vertical intercept. Factor the differences of these two conditions before making any assumptions about symmetry.
Solution
The tangent at \(a\) has slope \(4a^3-4a\) and intercept
\[f(a)-af^{\prime}(a)=-3a^4+2a^2.\]Equal slopes at distinct inputs give
\[4(a-b)(a^2+ab+b^2-1)=0, \qquad a^2+ab+b^2=1.\]Equal intercepts give
\[(a^2-b^2)\bigl[2-3(a^2+b^2)\bigr]=0.\]There are two cases.
If \(a^2=b^2\), distinctness forces \(b=-a\). The slope condition becomes \(a^2=1\), giving the pair \(-1,1\). Both slopes are zero and both heights are \(-1\), so the common tangent is
\[y=-1,\]at \((-1,-1)\) and \((1,-1)\).
Otherwise, the intercept condition forces \(a^2+b^2=2/3\). The slope condition then gives \(ab=1/3\), so
\[(a-b)^2=a^2+b^2-2ab=\frac23-\frac23=0.\]This contradicts distinctness. 
Hence there are no other common tangents. Symmetry suggests the answer, but the two conditions are what establish completeness.
Problem 13
Define
\[f(x)=\begin{cases}x^2\sin(1/x),&x\ne0,\\0,&x=0.\end{cases}\]- Prove from the definition that \(f^{\prime}(0)=0\).
- For every positive integer \(n\), put \(a=1/(n\pi)\). Use the derivative definition to show that \(f^{\prime}(a)=(-1)^{n+1}\).
- Can the tangent slopes at nearby inputs be forced close to the slope at zero just because the derivative exists at zero?
Do not assume a formula for the derivative of sine. Use the trigonometric limits from Module 1.
Hint 1
| At zero, bound the difference quotient by $$ | h | \(. At\)a=1/(n\pi)\(, write\)1/(a+h)=n\pi+u\(and compute\)u/h$$ exactly. |
Hint 2
Substitute \(\sin(n\pi+u)=(-1)^n\sin u\), then insert the factor \(\sin u/u\). The remaining factors have ordinary finite limits.
Solution
At zero,
\[f^{\prime}(0)=\lim_{h\to0}\frac{h^2\sin(1/h)}h =\lim_{h\to0}h\sin(1/h)=0,\]| by the squeeze theorem, since $$ | h\sin(1/h) | \le | h | $$. |
For \(a=1/(n\pi)\), we have \(f(a)=0\). Write
\[u=\frac1{a+h}-\frac1a=-\frac{h}{a(a+h)}.\]As \(h\to0\), \(u\to0\). The addition identity then gives
\[\begin{aligned} f^{\prime}(a) &=\lim_{h\to0}\frac{(a+h)^2\sin(n\pi+u)}h\\ &=(-1)^n\lim_{h\to0}(a+h)^2\frac{\sin u}{u}\frac{-1}{a(a+h)}\\ &=(-1)^{n+1}. \end{aligned}\]The inputs \(1/(n\pi)\) can be made arbitrarily close to zero, yet their tangent slopes alternate between \(1\) and \(-1\), rather than approaching \(f^{\prime}(0)=0\).
A derivative at a point measures secants anchored at that point. It does not, by itself, force nearby tangent slopes to be close to it.
Problem 14
Design a polynomial position model \(p(t)\) for a camera slider, measured in metres, during \(0\le t\le1\) second. It must satisfy
\[\begin{aligned} p(0)&=0,&p(1)&=1,\\ p^{\prime}(0)&=0,&p^{\prime}(1)&=0,\\ p^{\prime\prime}(0)&=0,&p^{\prime\prime}(1)&=0. \end{aligned}\]The camera must also be at position \(1/3\) at \(t=1/2\), and must never move backwards.
Find a polynomial of the smallest possible degree that meets all these conditions. Prove both that it works and that no lower degree can work. You may use the fact that a polynomial with a double zero at \(a\) has a factor \((t-a)^2\).
Hint 1
Start without the midpoint condition. The endpoint velocity and acceleration conditions force \(p^{\prime}(t)\) to have double zeros at both ends. Find the unique polynomial of degree at most five meeting the six endpoint conditions.
Hint 2
A correction preserving the value, first derivative, and second derivative at both ends has the factor \(t^3(1-t)^3\). Try a constant multiplier first. If that forces backward motion, try a linear multiplier. Factor the velocity before choosing any remaining coefficient.
Solution
First establish a lower bound. The velocity has double zeros at \(0\) and \(1\), so it contains \(t^2(1-t)^2\) as a factor. It is not identically zero, since the endpoint positions differ. Its degree is at least four; therefore the position degree is at least five.
A polynomial of degree at most five with the initial three conditions has the form \(At^3+Bt^4+Ct^5\). The final three conditions give
\[\begin{aligned} A+B+C&=1,\\ 3A+4B+5C&=0,\\ 6A+12B+20C&=0. \end{aligned}\]Solving gives the unique candidate
\[p_5(t)=10t^3-15t^4+6t^5.\]But \(p_5(1/2)=1/2\), so degree five cannot meet the midpoint requirement.
Rule out degree six. The difference between any other candidate and \(p_5\) must have its value and first two derivatives zero at each endpoint. For a polynomial, vanishing value and first two derivatives force the constant, linear, and quadratic coefficients in powers of the displacement from that endpoint to vanish. Each endpoint therefore contributes a triple factor, so the difference is divisible by \(t^3(1-t)^3\). For degree at most six, every candidate has the form
\[p_6(t)=p_5(t)+c\,t^3(1-t)^3.\]At the midpoint, \(1/2+c/64=1/3\) forces \(c=-32/3\). Its velocity is
\[p_6^{\prime}(t)=t^2(1-t)^2(64t-2),\]which is negative for \(0<t<1/32\). It initially moves backwards. Thus no polynomial of degree at most six can satisfy all the requirements.
Construct a degree-seven solution. Write
\[p(t)=p_5(t)+t^3(1-t)^3(a+bt).\]The midpoint condition is \(a+b/2=-32/3\). Differentiating and factoring gives
\[p^{\prime}(t)=t^2(1-t)^2\bigl[30+3a(1-2t)+b(4t-7t^2)\bigr].\]Choose \(a=-10\) so the bracket’s constant term is zero. The midpoint condition then gives \(b=-4/3\). This yields
\[\boxed{p(t)=10t^3-15t^4+6t^5-t^3(1-t)^3\left(10+\frac43t\right)}.\]The correction preserves all six endpoint conditions, and at the midpoint it changes \(1/2\) by \(-1/6\), giving \(1/3\). Its velocity simplifies to
\[p^{\prime}(t)=\frac43t^3(1-t)^2(41+7t)\ge0\qquad(0\le t\le1).\]It therefore never moves backwards. The degree is seven, and the exclusions above prove that seven is the smallest possible degree. The solution need not be unique among degree-seven polynomials; one verified construction suffices.