3 Problems
The rules of differentiation
The problems follow the sections of the theory. Those that are checked at once get new numbers each time you press New variant: repeat them until the rule runs by itself. Those with a solution are for paper. Write the whole solution before you open it, then compare line by line. Last come problems in the form of the exam, with points and a marking guide.
Typing an answer. Put a denominator with more than one factor in brackets: 3/(2 sqrt(x)), 5/(2x+1)^2, -8/x^3. The preview under the box shows how your answer was read. Look at it before you press Check.
Three problems look ahead. The ones with $e^{x}$ or $\ln x$ use two derivatives that Lecture 4 proves, $(e^{x})'=e^{x}$ and $(\ln x)'=1/x.$ They are stated at the end of the section on the chain rule.
●○○ Basic●●○ Intermediate●●● Challenge
Checked instantlySolution included
Limits
Read the theory: 1. Limits →For each limit, say first whether substituting works, and why. Then find the limit.
(a) $\displaystyle\lim_{x\to3}\frac{x^{2}-4x+1}{x+1}$
(b) $\displaystyle\lim_{x\to-1}\frac{x^{2}-1}{x+1}$
(c) $\displaystyle\lim_{x\to4}\frac{x^{2}-16}{x^{2}-5x+4}$
(d) $\displaystyle\lim_{h\to0}\frac{(2+h)^{3}-8}{h}.$ Which derivative is this limit?
Hint
Substituting works when the denominator is not zero at the point (Fact 3.4). When both the numerator and the denominator are zero there, both have the same factor: find it and cancel it. In (d), expand the cube.
(a) Substituting works. The function is rational and its denominator is $3+1=4\ne0$ at $x=3,$ so by Fact 3.4 the limit is the value: $$\frac{9-12+1}{4}=-\frac12\,.$$
(b) Substituting gives $\tfrac00.$ That is a signal, not an answer. The numerator is zero at $x=-1,$ so it has the factor $x+1{:}$ $$\begin{aligned} \frac{x^{2}-1}{x+1}&=\frac{(x-1)(x+1)}{x+1} \\ &=x-1 \end{aligned}$$ for $x\ne-1.$ The limit is $-1-1=-2.$
(c) Again $\tfrac00.$ Both polynomials are zero at $x=4,$ so both have the factor $x-4{:}$ $$\begin{aligned} \frac{x^{2}-16}{x^{2}-5x+4}&=\frac{(x-4)(x+4)}{(x-4)(x-1)} \\ &=\frac{x+4}{x-1} \end{aligned}$$ for $x\ne4.$ Now the denominator is $3\ne0$ at $x=4,$ and the limit is $\dfrac{8}{3}.$
(d) Again $\tfrac00,$ at $h=0.$ Expand: $(2+h)^{3}=8+12h+6h^{2}+h^{3}.$ So $$\begin{aligned} \frac{(2+h)^{3}-8}{h}&=\frac{12h+6h^{2}+h^{3}}{h} \\ &=12+6h+h^{2} \end{aligned}$$ for $h\ne0,$ and the limit is $12.$
Which derivative. The quotient is the Newton quotient of $f(x)=x^{3}$ at $a=2,$ so the limit is $f'(2).$ The power rule agrees: $f'(x)=3x^{2}$ and $f'(2)=12.$
Answer
(a) $-\tfrac12$ (substitution works).
(b) $-2.$
(c) $\tfrac83.$
(d) $12,$ which is $f'(2)$ for $f(x)=x^{3}.$
How did it go?
Stuck? Read 1. Limits →
The rules that need no thought
Read the theory: 2. The rules that need no thought →Each line contains one error. Say what went wrong, in words, and give the right derivative.
(a) $\left(\dfrac{4}{x^{2}}\right)'=-\dfrac{8}{x}$
(b) $\left(\dfrac{x^{5}}{5}+7\right)'=x^{4}+7$
(c) $\bigl(6\sqrt{x}\bigr)'=3\sqrt{x}$
Hint
Rewrite each function as a constant times a power of $x$ before you look at the derivative. Then follow the exponent: what was it, and what should it become?
(a) One was subtracted from the size of the exponent. $\dfrac{4}{x^{2}}=4x^{-2}.$ The exponent goes $-2\mapsto-3,$ not $-2\mapsto-1$ (Watch out 3.6): $$\left(4x^{-2}\right)'=4\cdot(-2)x^{-3}=-\frac{8}{x^{3}}\,.$$
(b) The additive constant was carried along. Additive constants disappear (Fact 3.5). The first term is right: $\tfrac15x^{5}$ is a constant multiple, and $\tfrac15\cdot5x^{4}=x^{4}.$ So the derivative is $x^{4}.$
(c) The exponent was brought down but not lowered. $6\sqrt{x}=6x^{1/2},$ and the exponent goes $\tfrac12\mapsto-\tfrac12{:}$ $$\left(6x^{1/2}\right)'=6\cdot\tfrac12x^{-1/2}=\frac{3}{\sqrt{x}}\,.$$
Answer
(a) $-\dfrac{8}{x^{3}}$
(b) $x^{4}$
(c) $\dfrac{3}{\sqrt{x}}$
How did it go?
Stuck? Read 2. The rules that need no thought →
Products and quotients
Read the theory: 3. Products and quotients →At the price $P$ kronor a kiosk sells $D(P)=120-4P$ cups of coffee a day, for $0\le P\le30.$ Its revenue is $R(P)=P\cdot D(P).$
(a) Find $R'(P)$ with the product rule, without multiplying out first.
(b) At $P=10,$ say in kronor what each of the two terms of $R'(10)$ means. Does a small price rise raise the revenue?
(c) The same at $P=20.$
(d) For which $P$ is $R'(P)=0?$ What is the sign of $R'$ on each side, and what is the largest revenue?
Hint
The two factors are $P$ and $120-4P,$ and both vary. In (b) and (c): one term is what you gain on the cups you still sell, the other is what you lose on the cups you no longer sell.
(a) The product rule. The factor $P$ has derivative $1$ and the factor $120-4P$ has derivative $-4{:}$ $$\begin{aligned} R'(P)&=1\cdot(120-4P) \\ &\quad+P\cdot(-4) \\ &=120-8P\,. \end{aligned}$$ (Multiplying out gives $R(P)=120P-4P^{2}$ and the same derivative.)
(b) At $P=10.$ The first term is $D(10)=80{:}$ a krona more on each of the $80$ cups sold. The second is $10\cdot(-4)=-40{:}$ four customers fewer for each krona, each of whom paid $10$ kronor. Together $R'(10)=80-40=40>0,$ so a small price rise raises the revenue, by about $40$ kronor a day per krona.
(c) At $P=20.$ Now $D(20)=40,$ and the loss is $20\cdot(-4)=-80.$ So $R'(20)=40-80=-40<0{:}$ fewer cups to gain on, and each lost customer costs more. A price rise lowers the revenue.
(d) Where the two terms balance. $R'(P)=120-8P=8(15-P)$ is zero for $P=15.$
P < 15 15 P > 15 ${R'(P)}$ + 0 − So $R$ is increasing on $[0,15]$ and decreasing on $[15,30]$ (Fact 2.23), and the largest revenue is $R(15)=15\cdot60=900$ kronor a day.
Answer
(a) $R'(P)=120-8P.$
(b) $R'(10)=80-40=40{:}$ revenue rises.
(c) $R'(20)=40-80=-40{:}$ revenue falls.
(d) $R'(P)=0$ at $P=15,$ with $R'>0$ before and $R'<0$ after. The largest revenue is $R(15)=900$ kronor a day.
How did it go?
Stuck? Read 3. Products and quotients →
The theory says that the quotient rule extends the power rule to negative integers. Here is that step.
(a) Let $g$ be differentiable with $g(x)\ne0,$ and take for granted that $\dfrac1g$ is differentiable too. Differentiate both sides of $g\cdot\dfrac1g=1$ with the product rule, and show that $$\left(\frac1g\right)'=-\frac{g'}{g^{2}}\,.$$
(b) Let $n$ be a natural number. Use (a) and the power rule for $x^{n}$ to show that $\bigl(x^{-n}\bigr)'=-n\,x^{-n-1}$ for $x\ne0.$
(c) Check that the formula in (a) is what the quotient rule (3.2) gives when the numerator is the constant $1.$
Hint
(a) The right-hand side is a constant, so its derivative is $0.$ Solve the equation you get for $\bigl(\tfrac1g\bigr)'.$
(b) $x^{-n}=\dfrac{1}{x^{n}},$ so take $g(x)=x^{n}.$
(a) Differentiate both sides. The left-hand side is a product, the right-hand side a constant: $$g'\cdot\frac1g+g\cdot\left(\frac1g\right)'=0\,.$$
(a) Solve. Move the first term over and divide by $g,$ which is allowed since $g(x)\ne0{:}$ $$\begin{aligned} g\cdot\left(\frac1g\right)'&=-\frac{g'}{g}\,,\quad\text{so} \\ \left(\frac1g\right)'&=-\frac{g'}{g^{2}}\,. \end{aligned}$$
(b) Take $g(x)=x^{n}.$ Then $g'(x)=n\,x^{n-1}$ by the power rule for a natural number, $g(x)\ne0$ for $x\ne0,$ and $g(x)^{2}=x^{2n}.$ So by (a) $$\begin{aligned} \bigl(x^{-n}\bigr)'&=-\frac{n\,x^{n-1}}{x^{2n}} \\ &=-n\,x^{n-1-2n} \\ &=-n\,x^{-n-1}\,. \end{aligned}$$ This is $a\,x^{a-1}$ with $a=-n{:}$ the power rule holds for negative integers, with the exponent going from $-n$ to $-n-1.$
(c) The quotient rule with $f\equiv1.$ Then $f'=0,$ and (3.2) gives $$\left(\frac1g\right)'=\frac{0\cdot g-1\cdot g'}{g^{2}}=-\frac{g'}{g^{2}}\,,$$ the same formula.
Answer
(a) $g'\cdot\tfrac1g+g\cdot\bigl(\tfrac1g\bigr)'=0,$ which gives $\bigl(\tfrac1g\bigr)'=-g'/g^{2}.$
(b) With $g=x^{n}{:}$ $\bigl(x^{-n}\bigr)'=-n\,x^{n-1}/x^{2n}=-n\,x^{-n-1}.$
(c) (3.2) with $f\equiv1,$ $f'=0$ gives $-g'/g^{2}.$
How did it go?
Stuck? Read 3. Products and quotients →
Relative rates: growth in per cent
Read the theory: 4. Relative rates: growth in per cent →A country's GDP is $Y$ and its population is $N,$ so GDP per head is $y=Y/N.$
(a) GDP grows by $3\%$ a year and the population by $1\%$ a year. About how fast does GDP per head grow? Compare with the exact figure when the rates are read as year-on-year factors.
(b) Derive the rule you used in (a): divide the quotient rule (3.2) by $f/g$ and simplify.
(c) GDP per head is to grow by $2\%$ a year while the population grows by $1.5\%$ a year. About how fast must GDP grow?
Hint
(a) A quotient: Fact 3.15. The year-on-year factors are $1.03$ and $1.01.$
(c) Write $Y=y\cdot N.$ Now it is a product.
(a) A quotient, so the relative rates subtract. $$\begin{aligned} \frac{y'}{y}&=\frac{Y'}{Y}-\frac{N'}{N} \\ &=3\%-1\%=2\%\,. \end{aligned}$$ As year-on-year factors: $1.03/1.01=1.0198$ to four decimals, a growth of $1.98\%.$ The difference $2\%$ is the approximation.
(b) Divide (3.2) by $f/g.$ Dividing by $f/g$ is multiplying by $g/f{:}$ $$\begin{aligned} \frac{(f/g)'}{f/g}&=\frac{f'g-fg'}{g^{2}}\cdot\frac{g}{f} \\ &=\frac{f'g-fg'}{fg} \\ &=\frac{f'}{f}-\frac{g'}{g}\,. \end{aligned}$$ In the last step the fraction is split in two, and $g$ cancels in the first part and $f$ in the second.
(c) Turn it round. $Y=y\cdot N$ is a product, so the relative rates add: $$\begin{aligned} \frac{Y'}{Y}&=\frac{y'}{y}+\frac{N'}{N} \\ &=2\%+1.5\%=3.5\%\,. \end{aligned}$$ As year-on-year factors: $1.02\times1.015=1.0353,$ a growth of $3.53\%.$
Answer
(a) About $2\%$ a year (exactly $1.98\%$ as year-on-year factors).
(b) $\dfrac{f'g-fg'}{g^{2}}\cdot\dfrac{g}{f}=\dfrac{f'}{f}-\dfrac{g'}{g}.$
(c) About $3.5\%$ a year.
How did it go?
Stuck? Read 4. Relative rates: growth in per cent →
The chain rule
Read the theory: 5. The chain rule →Differentiate with Method 3.20. For each function, say first which operation is outermost, and so which rule comes first.
(a) $f(x)=x^{2}(1-3x)^{4}$
(b) $f(x)=\left(\dfrac{2x+1}{x-1}\right)^{3}$ for $x\ne1$
(c) $f(x)=\dfrac{1}{(x^{2}+4)^{3}}$
Hint
The outermost operation is the last thing you would do when computing the value at a number. In (c), rewrite first: a constant over a power.
(a) Outermost: a product. The factors are $x^{2}$ and $(1-3x)^{4}.$ The second is a composition, with derivative $4(1-3x)^{3}\cdot(-3)=-12(1-3x)^{3}$ by the chain rule. So $$\begin{aligned} f'(x) &= 2x(1-3x)^{4} \\ &\quad -12x^{2}(1-3x)^{3} \\ &= 2x(1-3x)^{3} \\ &\qquad\cdot(1-3x-6x) \\ &= 2x(1-3x)^{3}(1-9x)\,. \end{aligned}$$
(b) Outermost: the third power. A composition with outside $y^{3}$ and inside $y=\dfrac{2x+1}{x-1}.$ The inside is a quotient: $$\begin{aligned} y'&=\frac{2(x-1)-(2x+1)\cdot1}{(x-1)^{2}} \\ &=\frac{-3}{(x-1)^{2}}\,. \end{aligned}$$ So $$\begin{aligned} f'(x)&=3\left(\frac{2x+1}{x-1}\right)^{2}\cdot\frac{-3}{(x-1)^{2}} \\ &=-\frac{9(2x+1)^{2}}{(x-1)^{4}}\,. \end{aligned}$$
(c) Rewrite first. $f(x)=(x^{2}+4)^{-3},$ and no quotient rule is needed. Outermost: the power $-3,$ a composition with inside $x^{2}+4{:}$ $$\begin{aligned} f'(x)&=-3(x^{2}+4)^{-4}\cdot2x \\ &=-\frac{6x}{(x^{2}+4)^{4}}\,. \end{aligned}$$
Check. In each case the outer derivative is evaluated at the inner value: $(1-3x),$ $\dfrac{2x+1}{x-1}$ and $x^{2}+4$ stay inside their brackets. In (c) the sign fits: for $x>0$ the denominator grows, so $f$ falls.
Answer
(a) Product rule first: $f'(x)=2x(1-3x)^{3}(1-9x).$
(b) Chain rule first: $f'(x)=-\dfrac{9(2x+1)^{2}}{(x-1)^{4}}.$
(c) Rewrite, then chain rule: $f'(x)=-\dfrac{6x}{(x^{2}+4)^{4}}.$
How did it go?
Stuck? Read 5. The chain rule →
Higher-order derivatives
Read the theory: 6. Higher-order derivatives →Let $f(x)=x^{3}-6x^{2}+9x+1.$
(a) Find $f'(x),$ factor it, and make a sign table. Where is $f$ increasing and where is it decreasing?
(b) Find $f''(x).$ Where does the graph bend upwards, and where downwards?
(c) Describe the graph in words, from left to right, using the points with $x=1,$ $x=2$ and $x=3.$
Hint
(a) is Method 2.25 of Lecture 2. (b) is Fact 3.22. In (c), say for each stretch both which way the curve goes and which way it bends.
(a) The first derivative. $f'(x)=3x^{2}-12x+9=3(x-1)(x-3),$ which is zero at $x=1$ and $x=3.$
x < 1 1 1 < x < 3 3 x > 3 ${x-1}$ − 0 + + + ${x-3}$ − − − 0 + ${f'(x)}$ + 0 − 0 + So $f$ is increasing on $(-\infty,1],$ decreasing on $[1,3]$ and increasing on $[3,\infty).$
(b) The second derivative. $f''(x)=6x-12=6(x-2).$ It is negative for $x<2$ and positive for $x>2.$ By Fact 3.22 the graph bends downwards for $x<2$ and upwards for $x>2.$
(c) Three points. $f(1)=5,$ $f(2)=3$ and $f(3)=1,$ and the slope at $x=2$ is $f'(2)=-3.$
(c) In words. The curve rises, bending downwards, to the point $(1,5),$ where its tangent is horizontal. It then falls. Between $x=1$ and $x=2$ it still bends downwards, so it falls more and more steeply, down to the slope $-3$ at $(2,3).$ There the bending changes. Between $x=2$ and $x=3$ it falls while bending upwards, so it flattens out, to a horizontal tangent at $(3,1).$ After that it rises, bending upwards, more and more steeply.
Answer
(a) $f'(x)=3(x-1)(x-3){:}$ increasing on $(-\infty,1]$ and on $[3,\infty),$ decreasing on $[1,3].$
(b) $f''(x)=6(x-2){:}$ bends downwards for $x<2,$ upwards for $x>2.$
(c) Up to $(1,5),$ down through $(2,3),$ where the slope is $-3$ and the bending changes, to $(3,1),$ then up again.
How did it go?
Stuck? Read 6. Higher-order derivatives →
Mixed, as on the exam
Here the sections are mixed, and nothing says which rule to use. Write the solution on paper, with reasons, before you look.
(a) At its present output a factory has a marginal cost of $40$ kronor per unit, and its output is rising by $15$ units per week. How fast is its total cost rising? Write the units of every number.
(b) Some months later the output is still rising by $15$ units per week, and the total cost by $900$ kronor per week. What is the marginal cost now?
(c) Another factory has the cost $C(Q)=0.5Q^{2}+10Q$ kronor for $Q$ units, and produces $Q(t)=20+3t$ units in week $t.$ Find how fast its cost is rising in week $t=4$ with the chain rule. Then check by substituting $Q(t)$ into $C$ first.
Hint
Write $\dfrac{dC}{dt}=\dfrac{dC}{dQ}\cdot\dfrac{dQ}{dt}$ with the units under each factor. In (c), at which value of $Q$ is $\dfrac{dC}{dQ}$ to be evaluated?
(a) Rates multiply. $$\begin{aligned} \frac{dC}{dt}&=\frac{dC}{dQ}\cdot\frac{dQ}{dt} \\ &=40\ \frac{\text{kr}}{\text{unit}}\times15\ \frac{\text{units}}{\text{week}} \\ &=600\ \frac{\text{kr}}{\text{week}}\,. \end{aligned}$$ The units cancel to kronor per week, which is what a rate of cost over time must have.
(b) The same equation, solved for the other factor. $900=\dfrac{dC}{dQ}\cdot15,$ so $$\begin{aligned} \frac{dC}{dQ}&=\frac{900\ \text{kr/week}}{15\ \text{units/week}} \\ &=60\ \text{kr per unit}\,. \end{aligned}$$ Here dividing is right, and again the units say so.
(c) The chain rule. $C'(Q)=Q+10$ and $Q'(t)=3.$ In week $4$ the output is $Q(4)=32,$ and the outer derivative is evaluated there: $$\begin{aligned} \frac{dC}{dt}&=C'\bigl(Q(4)\bigr)\cdot Q'(4) \\ &=(32+10)\cdot3 \\ &=126\ \text{kr per week}\,. \end{aligned}$$ Evaluating $C'$ at $4$ instead of at $32$ would give $14\cdot3=42,$ which is Watch out 3.17.
(c) The check. Substituting first, $$\begin{aligned} C\bigl(Q(t)\bigr)&=0.5(20+3t)^{2} \\ &\quad+10(20+3t)\,, \end{aligned}$$ with derivative $0.5\cdot2(20+3t)\cdot3+30=3(20+3t)+30.$ At $t=4$ this is $96+30=126.$
Answer
(a) $600$ kronor per week.
(b) $60$ kronor per unit.
(c) $126$ kronor per week.
How did it go?
Stuck? Read 5. The chain rule →
(a) Let $f$ be twice differentiable on an interval $I,$ with $f''(x)>0$ for every $x$ in $I,$ and suppose that $f'(c)=0$ for some point $c$ in $I.$ Show that $f(x)\ge f(c)$ for every $x$ in $I{:}$ the value $f(c)$ is the smallest that $f$ takes on $I.$
(b) Let $f(x)=x^{2}+\dfrac{16}{x}$ for $x>0.$ Use (a) to find the smallest value of $f.$
(c) Why does (a) say nothing about this $f$ for $x<0?$
Hint
(a) Use Fact 2.23 twice: first on $f',$ whose derivative is $f'',$ and then on $f.$ What is the sign of $f'$ to the left of $c,$ and to the right?
(b) Rewrite $\dfrac{16}{x}$ as $16x^{-1}$ before you differentiate.
(a) The slope is increasing. $f''$ is the derivative of $f',$ and it is positive on $I.$ By Fact 2.23, used on $f',$ the function $f'$ is strictly increasing on $I.$
(a) So the slope changes sign at $c.$ Since $f'(c)=0$ and $f'$ is strictly increasing, $f'(x)<0$ for $x<c$ and $f'(x)>0$ for $x>c$ (for $x$ in $I$).
(a) So $f$ falls, then rises. On the part of $I$ to the left of $c,$ with $c$ included, $f'\le0,$ so $f$ is decreasing there by Fact 2.23: $f(x)\ge f(c)$ for $x<c.$ On the part to the right, $f'\ge0,$ so $f$ is increasing: $f(x)\ge f(c)$ for $x>c.$ Hence $f(x)\ge f(c)$ on all of $I.$
(b) The derivatives. $f(x)=x^{2}+16x^{-1},$ so $$\begin{aligned} f'(x)&=2x-\frac{16}{x^{2}}\,, \\ f''(x)&=2+\frac{32}{x^{3}}\,. \end{aligned}$$ For $x>0$ both terms of $f''$ are positive, so $f''>0$ on the interval $x>0.$
(b) Where $f'$ is zero. The equation $2x=\dfrac{16}{x^{2}}$ gives $x^{3}=8,$ so $x=2,$ and indeed $f'(2)=4-4=0.$ By (a) the smallest value of $f$ for $x>0$ is $$f(2)=4+8=12\,.$$
(c) The function is not defined at $x=0,$ so $x<0$ and $x>0$ are two separate intervals and (a) has to be used on each by itself. For $x<0$ there is no point $c$ to start from: both $2x$ and $-\dfrac{16}{x^{2}}$ are negative, so $f'(x)<0$ throughout.
Answer
(a) $f''>0$ makes $f'$ strictly increasing, so $f'<0$ before $c$ and $f'>0$ after it: $f$ decreases to $c$ and increases after it.
(b) $f'(x)=2x-16/x^{2}=0$ at $x=2,$ and $f''>0$ for $x>0.$ The smallest value is $f(2)=12.$
(c) For $x<0$ the derivative is negative everywhere, so there is no point with $f'(c)=0.$
How did it go?
Stuck? Read 6. Higher-order derivatives →
As on the exam
3 problems · 15 pProblems in exam form, with points. Solve them on paper or in the box. Then open the solution one step at a time and mark yourself against the rubric.
A rational function: derivative, sign table, increase and decrease
Let $$f(x)=\frac{x^{2}-3x+6}{x-2}\,.$$
- (a)3 p
State the domain of $f.$ Find $f'(x),$ and find every $x$ with $f'(x)=0.$
- (b)3 p
Make a sign table for $f',$ and give the intervals on which $f$ is increasing and the intervals on which it is decreasing.
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(a) The domain. The denominator is zero only at $x=2,$ so $f$ is defined for all $x\ne2.$ (The numerator is $4-6+6=4\ne0$ there, so nothing cancels.)
(a) The quotient rule. The numerator $x^{2}-3x+6$ has derivative $2x-3,$ and the denominator $x-2$ has derivative $1.$ By (3.2), $f'(x)=\dfrac{N(x)}{(x-2)^{2}}$ with the numerator $$\begin{aligned} N(x)&=(2x-3)(x-2) \\ &\quad-(x^{2}-3x+6)\cdot1\,. \end{aligned}$$
(a) Simplify the numerator. $(2x-3)(x-2)=2x^{2}-7x+6,$ and the whole second bracket is subtracted: $$\begin{aligned} N(x)&=2x^{2}-7x+6 \\ &\quad-x^{2}+3x-6 \\ &=x^{2}-4x=x(x-4)\,. \end{aligned}$$ So $$f'(x)=\frac{x(x-4)}{(x-2)^{2}}\qquad(x\ne2)\,.$$
(a) Where $f'(x)=0.$ A quotient is zero exactly where its numerator is zero and the quotient is defined: $x=0$ and $x=4.$ Both are in the domain.
(b) The sign table. The denominator $(x-2)^{2}$ is positive for every $x\ne2,$ so the sign of $f'$ is the sign of $x(x-4).$ The point $x=2,$ where $f$ and $f'$ are not defined, gets a column of its own.
x < 0 0 0 < x < 2 2 2 < x < 4 4 x > 4 $x$ − 0 + + + + + ${x-4}$ − − − − − 0 + ${(x-2)^{2}}$ + + + 0 + + + ${f'(x)}$ + 0 − not defined − 0 + (b) The intervals. By the sign of the derivative (Fact 2.23, Method 2.25), on each interval inside the domain:
$f$ is increasing on $(-\infty,0]$ and on $[4,\infty);$
$f$ is decreasing on $[0,2)$ and on $(2,4].$
The two intervals of decrease must be stated separately. $f$ is not decreasing on the set from $0$ to $4$ as a whole: $f(1)=-4$ and $f(3)=6,$ so the values go up across the gap at $x=2.$
Answer
(a) Domain: all $x\ne2.$ $f'(x)=\dfrac{x(x-4)}{(x-2)^{2}},$ and $f'(x)=0$ for $x=0$ and $x=4.$
(b) Increasing on $(-\infty,0]$ and on $[4,\infty).$ Decreasing on $[0,2)$ and on $(2,4].$
Mark your solution
Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
Your points 0 of 6 p
Common errors
- The numerator of the quotient rule in the wrong order, which gives $f'(x)=-x(x-4)/(x-2)^{2}$ and every sign in the table reversed. Put $g\equiv1$ in the formula you wrote: it must give $f',$ not $-f'$ (Watch out 3.12). In the practice problems on the quotient rule fewer than half of the answers were right.
- Subtracting only the first term of the second bracket: $2x^{2}-7x+6-x^{2}-3x+6=x^{2}-10x+12.$ The whole of $x^{2}-3x+6$ is subtracted, so write the bracket.
- Leaving $x=2$ out of the sign table and answering “decreasing on $[0,4]$”. The function is not defined at $2,$ and $f(1)=-4<f(3)=6.$
- Giving $x=2$ as a point where $f'(x)=0.$ There the denominator is zero. The numerator $x(x-4)$ is $-4$ at $x=2,$ and $f'$ is not defined.
Differentiating with the letters left in
Let $a$ and $b$ be constants, and let $$f(x)=\frac{x^{2}+ax+b}{x-3}\,,\qquad x\ne3\,.$$
- (a)3 p
Find $f'(x),$ with the numerator simplified as far as possible.
- (b)1 p
What does the condition $f'(1)=0$ say about $a$ and $b?$ Does it determine them?
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(a) The two derivatives. $a$ and $b$ are constants, so $(x^{2}+ax+b)'=2x+a{:}$ the term $ax$ is a constant multiple of $x,$ and the additive constant $b$ disappears (Fact 3.5). The denominator $x-3$ has derivative $1.$
(a) The quotient rule. By (3.2), $f'(x)=\dfrac{N(x)}{(x-3)^{2}}$ with the numerator $$\begin{aligned} N(x)&=(2x+a)(x-3) \\ &\quad-(x^{2}+ax+b)\cdot1\,. \end{aligned}$$
(a) Simplify the numerator. $(2x+a)(x-3)=2x^{2}-6x+ax-3a,$ and the whole second bracket is subtracted: $$\begin{aligned} N(x)&=2x^{2}-6x+ax-3a \\ &\quad-x^{2}-ax-b \\ &=x^{2}-6x-3a-b\,. \end{aligned}$$ The two terms with $ax$ cancel. So $$f'(x)=\frac{x^{2}-6x-3a-b}{(x-3)^{2}}\,.$$
(b) Insert $x=1.$ The denominator is $(1-3)^{2}=4\ne0,$ so $f'(1)=0$ exactly when the numerator is zero at $x=1{:}$ $$\begin{aligned} 1-6-3a-b&=0\,,\quad\text{that is} \\ 3a+b&=-5\,. \end{aligned}$$
(b) One equation, two unknowns. The condition does not determine $a$ and $b{:}$ every pair with $b=-5-3a$ satisfies it, for instance $(a,b)=(0,-5)$ and $(a,b)=(1,-8).$ A second condition is needed to fix them, such as a given value of $f(1).$ (A check with $(a,b)=(0,-5){:}$ then $f'(x)=\dfrac{x^{2}-6x+5}{(x-3)^{2}}=\dfrac{(x-1)(x-5)}{(x-3)^{2}},$ which is zero at $x=1.$)
Answer
(a) $f'(x)=\dfrac{x^{2}-6x-3a-b}{(x-3)^{2}}.$
(b) $f'(1)=0$ exactly when $3a+b=-5.$ That is one equation for two unknowns, so it does not determine $a$ and $b.$
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Common errors
- Differentiating the letters: $(ax)'$ written as $a+x$ or as $ax,$ or $b$ kept in the derivative. $a$ and $b$ are constants. Only $x$ varies.
- Dropping the bracket around the subtracted numerator, which leaves $x^{2}-6x-3a+2ax+b$ or something like it. If your simplified numerator still contains $ax,$ look for this.
- Inserting $x=1$ into $f$ and not into $f'{:}$ the condition is about the derivative.
- Answering (b) with one pair of numbers. One equation in two unknowns has a whole line of solutions.
First and second derivative of a composition, and the shape of the graph
Let $$f(x)=\bigl(x^{2}-5\bigr)^{3}\,.$$
- (a)2 p
Find $f'(x)$ and $f'(2).$
- (b)2 p
Find $f''(x),$ written as a product of factors, and $f''(2).$
- (c)1 p
The point $x=2$ lies in the interval $1<x<\sqrt5.$ Is $f$ increasing on that interval? Which way does the graph bend there, and does the tangent at $x=2$ lie above or below the graph near $x=2?$
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(a) The chain rule. $f$ is a composition with outside $y^{3}$ and inside $y=x^{2}-5.$ The outer derivative $3y^{2}$ is evaluated at the inner value, and the inside has derivative $2x{:}$ $$\begin{aligned} f'(x)&=3\bigl(x^{2}-5\bigr)^{2}\cdot2x \\ &=6x\bigl(x^{2}-5\bigr)^{2}\,. \end{aligned}$$
(a) At $x=2.$ The inner value is $2^{2}-5=-1,$ so $$f'(2)=6\cdot2\cdot(-1)^{2}=12\,.$$
(b) The product rule, with the chain rule inside. $f'$ is the product of $6x,$ with derivative $6,$ and $\bigl(x^{2}-5\bigr)^{2},$ with derivative $2\bigl(x^{2}-5\bigr)\cdot2x=4x\bigl(x^{2}-5\bigr){:}$ $$\begin{aligned} f''(x)&=6\bigl(x^{2}-5\bigr)^{2} \\ &\quad+24x^{2}\bigl(x^{2}-5\bigr)\,. \end{aligned}$$
(b) Factor, and insert. Take out $6\bigl(x^{2}-5\bigr){:}$ $$\begin{aligned} f''(x) &= 6\bigl(x^{2}-5\bigr) \\ &\qquad\cdot\bigl(x^{2}-5+4x^{2}\bigr) \\ &= 6\bigl(x^{2}-5\bigr)\bigl(5x^{2}-5\bigr) \\ &= 30\bigl(x^{2}-5\bigr)\bigl(x^{2}-1\bigr)\,. \end{aligned}$$ At $x=2{:}$ $f''(2)=30\cdot(-1)\cdot3=-90.$ (The unfactored form agrees: $6\cdot1+24\cdot4\cdot(-1)=-90.$)
(c) The tangent. Where $f''<0$ the tangents lie above the graph (Fact 3.22). So the tangent at $x=2,$ which is $y=f(2)+f'(2)(x-2)=-1+12(x-2),$ lies above the graph near $x=2.$
Answer
(a) $f'(x)=6x\bigl(x^{2}-5\bigr)^{2}$ and $f'(2)=12.$
(b) $f''(x)=30\bigl(x^{2}-5\bigr)\bigl(x^{2}-1\bigr)$ and $f''(2)=-90.$
(c) On $1<x<\sqrt5$ we have $f'>0$ and $f''<0{:}$ $f$ is increasing, the graph bends downwards, and the tangent at $x=2$ lies above it.
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Tick each criterion that is really in your solution. A criterion scores only if what it asks for is there.
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Common errors
- The outer derivative evaluated at $x$ and not at the inner value: $3x^{2}\cdot2x.$ It looks respectable, and it is the error that Watch out 3.17 names.
- The inner derivative forgotten: $f'(x)=3\bigl(x^{2}-5\bigr)^{2},$ and then $f'(2)=3.$
- In $f'',$ only one factor of $6x\bigl(x^{2}-5\bigr)^{2}$ differentiated. In the live vote on a product, about a third of the answers differentiated one factor and kept the other as if it were a constant.
- Reading the bending from $f'{:}$ “$f'(2)=12>0,$ so the graph bends upwards”. Which way the graph goes is read from $f'.$ How it bends is read from $f''.$