4 Problems
Exponentials, logarithms, and approximation
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Three problems build the exam's implicit function step by step: the value $y(x_{0})$ from a cubic, then $y'(x_{0}),$ then $y''(x_{0}).$ The last one uses the addition in the section on implicit differentiation.
●○○ Basic●●○ Intermediate●●● Challenge
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The exponential function, and why $e$
Read the theory: 1. The exponential function, and why $e$ →For each function, say first whether the variable is in the base, in the exponent, or nowhere. Then differentiate (for $x>0$).
(a) $x^{4}$
(b) $4^{x}$
(c) $e^{4}$
(d) $x^{\pi}$
(e) $\pi^{x}$
Hint
Watch out 4.3: ask which slot the variable is in before asking which rule applies. A power function takes the power rule (Fact 4.14), an exponential takes Fact 4.4.
(a) Variable in the base. A power function with exponent $4{:}$ $\bigl(x^{4}\bigr)'=4x^{3}.$
(b) Variable in the exponent. An exponential with base $4.$ By Fact 4.4, $\bigl(4^{x}\bigr)'=4^{x}\ln4.$
(c) Variable nowhere. $e^{4}\approx54.6$ is a constant, so its derivative is $0.$ Neither $4e^{3}$ nor $e^{4}$ is right.
(d) Variable in the base. The exponent $\pi$ is a constant, however ugly, so by Fact 4.14 $\bigl(x^{\pi}\bigr)'=\pi x^{\pi-1}.$
(e) Variable in the exponent. An exponential with base $\pi{:}$ $\bigl(\pi^{x}\bigr)'=\pi^{x}\ln\pi.$
Answer
(a) $4x^{3}$
(b) $4^{x}\ln4$
(c) $0$
(d) $\pi x^{\pi-1}$
(e) $\pi^{x}\ln\pi$
How did it go?
Stuck? Read 1. The exponential function, and why $e$ →
The logarithm
Read the theory: 2. The logarithm →(a) Where is $f(x)=\ln\bigl(x^{2}-4x\bigr)$ defined? Find $f'(x).$ The formula for $f'$ is zero at $x=2.$ Does the graph of $f$ have a horizontal tangent there?
(b) Where is $g(x)=\ln\dfrac{x-1}{x+2}$ defined? Find $g'(x)$ with formula (4.2), and simplify.
(c) For $x>1$ one may first write $g(x)=\ln(x-1)-\ln(x+2).$ Check that this gives the same $g'.$ Why may this not be done for $x<-2?$
Hint
Watch out 4.6: the logarithm needs a positive argument. In (a) factor $x^{2}-4x;$ in (b) make a sign table for the quotient. Then use $\bigl(\ln h\bigr)'=h'/h.$
(a) The domain. $x^{2}-4x=x(x-4)$ is positive exactly when both factors have the same sign: for $x<0$ and for $x>4.$
(a) The derivative. With $h(x)=x^{2}-4x$ and $h'(x)=2x-4,$ $$ f'(x)=\frac{2x-4}{x^{2}-4x}\,, $$ for $x<0$ and for $x>4.$ The formula is zero at $x=2,$ but $2$ is not in the domain: there is no graph there, and no tangent. On the domain, $f'<0$ for $x<0$ and $f'>0$ for $x>4,$ so the graph has no horizontal tangent at all.
(b) The domain. The quotient $\dfrac{x-1}{x+2}$ is positive exactly when $x-1$ and $x+2$ have the same sign:
x < −2 −2 < x < 1 x > 1 ${x-1}$ − − + ${x+2}$ − + + quotient + − + So $g$ is defined for $x<-2$ and for $x>1.$
(b) The derivative. With $h(x)=\dfrac{x-1}{x+2},$ the quotient rule gives $$ \begin{aligned} h'(x)&=\frac{(x+2)-(x-1)}{(x+2)^{2}}\\ &=\frac{3}{(x+2)^{2}}\,, \end{aligned} $$ and by (4.2) $$ \begin{aligned} g'(x)&=\frac{3}{(x+2)^{2}}\cdot\frac{x+2}{x-1} \\ &=\frac{3}{(x+2)(x-1)}\,. \end{aligned} $$
(c) For $x>1.$ Both $x-1$ and $x+2$ are positive, so Fact 2.9 applies, and $$ \begin{aligned} g'(x)&=\frac{1}{x-1}-\frac{1}{x+2} \\ &=\frac{(x+2)-(x-1)}{(x-1)(x+2)} \\ &=\frac{3}{(x-1)(x+2)}\,, \end{aligned} $$ the same.
(c) For $x<-2.$ Now $x-1$ and $x+2$ are both negative, and $\ln(x-1)$ and $\ln(x+2)$ do not exist: Fact 2.9 is for positive numbers only. The formula from (b) is still right there, because (4.2) needs only $h(x)>0,$ and the quotient is positive.
Answer
(a) Defined for $x<0$ and $x>4;$ $f'(x)=\dfrac{2x-4}{x^{2}-4x}.$ No horizontal tangent: $x=2$ is outside the domain.
(b) Defined for $x<-2$ and $x>1;$ $g'(x)=\dfrac{3}{(x+2)(x-1)}.$
(c) The same $g'$ for $x>1.$ For $x<-2$ the two logarithms do not exist, though the formula from (b) still holds.
How did it go?
Stuck? Read 2. The logarithm →
Implicit differentiation
Read the theory: 3. Implicit differentiation →The curve $x^{2}+xy+y^{2}=12$ is an ellipse, tilted.
(a) Find $y'$ in terms of $x$ and $y.$
(b) At which points of the curve is the tangent horizontal?
(c) At which points does the formula for $y'$ break down, and what does the curve look like there?
Hint
Method 4.9; the term $xy$ needs the product rule. In (b), $y'=0$ gives a relation between $x$ and $y.$ Put it into the equation of the curve.
(a) Differentiate both sides. The product rule on $xy$ gives $y+xy'{:}$ $$ 2x+y+xy'+2y\,y'=0\,, $$ and solving the linear equation for $y',$ $$ y'=-\frac{2x+y}{x+2y}\,. $$
(b) Horizontal tangent: $y'=0.$ The numerator must be zero (with the denominator not zero): $y=-2x.$ On the curve, $$ x^{2}-2x^{2}+4x^{2}=3x^{2}=12\,, $$ so $x=\pm2.$ The points are $(2,-4)$ and $(-2,4).$ There $x+2y=-6$ and $6,$ not zero.
(c) The denominator is zero: $x=-2y.$ On the curve, $4y^{2}-2y^{2}+y^{2}=3y^{2}=12,$ so $y=\pm2{:}$ the points $(-4,2)$ and $(4,-2).$ There the numerator is $-6$ and $6,$ not zero, so the differentiated equation cannot be solved for $y'.$ The tangent is vertical: these are the leftmost and the rightmost points of the ellipse.
Answer
(a) $y'=-\dfrac{2x+y}{x+2y}.$
(b) At $(2,-4)$ and $(-2,4).$
(c) At $(-4,2)$ and $(4,-2),$ where the tangent is vertical.
How did it go?
Stuck? Read 3. Implicit differentiation →
The ellipse $x^{2}+4y^{2}=8$ passes through $(2,1).$
(a) Find $y'$ by implicit differentiation.
(b) Differentiate once more and show that, at every point of the ellipse with $y\ne0,$ $$ y''=-\frac{1}{2y^{3}}\,. $$
(c) Find $y'(2)$ and $y''(2)$ on the half through $(2,1).$ Which way do the upper and the lower half of the ellipse bend?
Hint
(b) is the addition after Example 4.11. Differentiate $2x+8y\,y'=0,$ solve for $y'',$ insert $y'$ from (a), and at the end use the equation of the ellipse itself.
(a) Once. $2x+8y\,y'=0,$ so $$ y'=-\frac{x}{4y}\,. $$
(b) Twice. $8y\,y'$ is a product of two functions of $x{:}$ $$ 2+8(y')^{2}+8y\,y''=0\,, $$ so $$ y''=-\frac{1+4(y')^{2}}{4y}\,. $$
(b) Insert $y'.$ $4(y')^{2}=\dfrac{x^{2}}{4y^{2}},$ so $$ \begin{aligned} 1+4(y')^{2}&=\frac{4y^{2}+x^{2}}{4y^{2}} \\ &=\frac{8}{4y^{2}}=\frac{2}{y^{2}}\,, \end{aligned} $$ by the equation of the ellipse. Hence $$ y''=-\frac{2}{y^{2}}\cdot\frac{1}{4y}=-\frac{1}{2y^{3}}\,. $$
(c) At $(2,1).$ $y'(2)=-\dfrac{2}{4}=-\dfrac12$ and $y''(2)=-\dfrac{1}{2}.$
(c) The two halves. On the upper half $y>0,$ so $y''<0{:}$ it bends downwards (Fact 3.22). On the lower half $y<0,$ so $y''>0{:}$ it bends upwards. Both halves bend away from the centre, as an ellipse should.
Answer
(a) $y'=-\dfrac{x}{4y}.$
(b) $y''=-\dfrac{1+4(y')^{2}}{4y}=-\dfrac{1}{2y^{3}}.$
(c) $y'(2)=-\tfrac12,$ $y''(2)=-\tfrac12.$ The upper half bends downwards, the lower half upwards.
How did it go?
Stuck? Read 3. Implicit differentiation →
Logarithmic differentiation, and the power rule in general
Read the theory: 4. Logarithmic differentiation, and the power rule in general →Let $y=x^{1/x}$ for $x>0.$
(a) Find $y'$ by logarithmic differentiation.
(b) Where is $y$ increasing and where is it decreasing? What is its largest value?
(c) Which is larger, $\sqrt{2}$ or $\sqrt[3]{3}?$ Decide with (b), without a calculator.
Hint
(a) $\ln y=\dfrac{\ln x}{x}.$ (b) Only one factor of $y'$ can change sign. (c) $\sqrt2=2^{1/2}=4^{1/4}.$ Where do $3$ and $4$ lie compared with $e?$
(a) Take logarithms. $\ln y=\dfrac{\ln x}{x}.$ Differentiate both sides, the right by the quotient rule: $$ \frac{y'}{y}=\frac{\frac1x\cdot x-\ln x}{x^{2}}=\frac{1-\ln x}{x^{2}}\,. $$
(a) Multiply back. $$ y'=x^{1/x}\cdot\frac{1-\ln x}{x^{2}}\,. $$
(b) The sign. $x^{1/x}>0$ and $x^{2}>0,$ so $y'$ has the sign of $1-\ln x{:}$ positive for $0<x<e,$ zero at $x=e,$ negative for $x>e.$
0 < x < e e x > e ${1-\ln x}$ + 0 − ${y'}$ + 0 − By Fact 2.23, $y$ is increasing on $(0,e]$ and decreasing on $[e,\infty).$ Its largest value is $y(e)=e^{1/e}\approx1.445.$
(c) Compare. $\sqrt2=2^{1/2}=4^{1/4}=y(4)$ and $\sqrt[3]{3}=3^{1/3}=y(3).$ Since $e<3<4$ and $y$ is strictly decreasing for $x\ge e,$ we get $y(3)>y(4){:}$ $\sqrt[3]{3}>\sqrt2.$ (Indeed $\sqrt[3]3\approx1.442$ and $\sqrt2\approx1.414.$)
Answer
(a) $y'=x^{1/x}\,\dfrac{1-\ln x}{x^{2}}.$
(b) Increasing on $(0,e],$ decreasing on $[e,\infty);$ largest value $e^{1/e}\approx1.445.$
(c) $\sqrt[3]{3}>\sqrt{2},$ because $\sqrt2=4^{1/4}$ and $y$ decreases from $x=3$ to $x=4.$
How did it go?
Stuck? Read 4. Logarithmic differentiation, and the power rule in general →
Linear approximation
Read the theory: 5. Linear approximation →(a) Use the linear approximation of $f(x)=\dfrac1x$ about a suitable point to estimate $\dfrac{1}{4.1}.$
(b) Use the linear approximation of $e^{x}$ about a suitable point to estimate $e^{-0.05}.$
(c) Without a calculator: is each estimate too high or too low? Give the reason.
Hint
(a) About $a=4.$ $f(4)=\tfrac14$ and $f'(x)=-\dfrac{1}{x^{2}},$ so $f'(4)=-\tfrac{1}{16}.$ With $x=4.1{:}$ $$ \begin{aligned} \frac{1}{4.1}&\approx\frac14-\frac{1}{16}\cdot0.1 \\ &=0.25-0.00625\\ &=0.24375\,. \end{aligned} $$
(b) About $a=0.$ $e^{0}=1$ and the derivative at $0$ is $1,$ so $e^{x}\approx1+x$ and $e^{-0.05}\approx0.95.$
(c) Which side of the curve the tangent lies on. $f''(x)=\dfrac{2}{x^{3}}>0$ for $x>0,$ and $(e^{x})''=e^{x}>0.$ Both graphs bend upwards, so by Fact 3.22 their tangents lie below them: both estimates are too low.
(c) Check. $\dfrac{1}{4.1}=0.24390\ldots>0.24375$ and $e^{-0.05}=0.95123\ldots>0.95.$
Answer
(a) $\dfrac{1}{4.1}\approx0.24375.$
(b) $e^{-0.05}\approx0.95.$
(c) Both are too low: both graphs bend upwards, so their tangents lie below them.
How did it go?
Stuck? Read 5. Linear approximation →
Taylor polynomials
Read the theory: 6. Taylor polynomials →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 firm produces $Q=\sqrt{K}\,\sqrt{L}$ units from $K$ units of capital and $L$ units of labour ($K,L>0$). Along the curve $\sqrt{K}\,\sqrt{L}=10$ the output is always $10,$ and $K$ is a function of $L.$
(a) Find $K$ when $L=4.$ Find $\dfrac{dK}{dL}$ by implicit differentiation, and its value at $L=4.$ What does the number say?
(b) Use the tangent line to estimate the capital needed for the same output when $L=4.2.$
(c) Find $\dfrac{d^{2}K}{dL^{2}}$ at $L=4$ by differentiating once more. Is the estimate in (b) too high or too low?
Hint
Write $K'$ for $\dfrac{dK}{dL}.$ The left-hand side is a product of $\sqrt K,$ whose derivative is $\dfrac{K'}{2\sqrt K}$ by the chain rule, and $\sqrt L.$
(a) The point. $\sqrt{4K}=10$ gives $4K=100,$ so $K=25.$
(a) Differentiate both sides with respect to $L.$ By the product rule and the chain rule, $$ \frac{K'}{2\sqrt K}\sqrt L+\sqrt K\,\frac{1}{2\sqrt L}=0\,. $$ Multiply by $2\sqrt K\sqrt L{:}$ $K'L+K=0,$ so $$ \begin{aligned} K'&=-\frac{K}{L}\,,\\ K'(4)&=-\frac{25}{4}=-6.25\,. \end{aligned} $$ One more unit of labour replaces about $6.25$ units of capital at this point, with the output unchanged.
(b) The tangent. By Fact 4.15 with $a=4{:}$ $$ \begin{aligned} K(4.2)&\approx25-6.25\cdot0.2 \\ &=23.75\,. \end{aligned} $$
(c) Once more. Differentiate $K'L+K=0{:}$ $K''L+K'+K'=0,$ so $K''=-\dfrac{2K'}{L}.$ At $L=4,$ $$ \begin{aligned} K''(4)&=-\frac{2\cdot(-6.25)}{4}\\ &=3.125>0\,. \end{aligned} $$
(c) The sign of the error. $K''>0,$ so the curve bends upwards near $L=4$ and its tangent lies below it (Fact 3.22): $23.75$ is too low. Indeed $K=100/L$ exactly, and $K(4.2)=23.81\ldots$
Answer
(a) $K=25;$ $\dfrac{dK}{dL}=-\dfrac{K}{L},$ which is $-6.25$ at $L=4{:}$ a unit of labour replaces about $6.25$ units of capital.
(b) $K(4.2)\approx23.75.$
(c) $K''(4)=3.125>0{:}$ the estimate is too low (exactly, $K(4.2)=23.81\ldots$).
How did it go?
Stuck? Read 3. Implicit differentiation →
Let $f(x)=x^{x}$ for $x>0.$
(a) Find $f'(x)$ and $f''(x).$
(b) Find the Taylor polynomial of degree $2$ of $f$ about $a=1.$
(c) Use it to estimate $1.1^{1.1}$ and $0.9^{0.9}.$
Hint
(a) Example 4.13 gives $f'.$ For $f''$ use the product rule on $x^{x}\cdot(\ln x+1),$ and $f'$ again for the first factor.
(a) The first derivative. By Example 4.13, $f'(x)=x^{x}(\ln x+1).$
(a) The second. The product rule, with $(x^{x})'=x^{x}(\ln x+1)$ and $(\ln x+1)'=\dfrac1x{:}$ $$ \begin{aligned} f''(x)&=x^{x}(\ln x+1)^{2} \\ &\quad+x^{x}\cdot\frac{1}{x}\,. \end{aligned} $$
(b) The numbers at $a=1.$ $f(1)=1,$ $f'(1)=1\cdot(0+1)=1$ and $f''(1)=1+1=2.$ By Definition 4.19, $$ \begin{aligned} P_{2}(x)&=1+(x-1)\\ &\quad+\frac{2}{2}(x-1)^{2} \\ &=1+(x-1)\\ &\quad+(x-1)^{2}\,. \end{aligned} $$
(c) Two estimates. At $x=1.1{:}$ $1.1^{1.1}\approx1+0.1+0.01=1.11.$ At $x=0.9{:}$ $0.9^{0.9}\approx1-0.1+0.01=0.91.$ (The true values are $1.1105\ldots$ and $0.9095\ldots$)
Answer
(a) $f'(x)=x^{x}(\ln x+1),$ $f''(x)=x^{x}(\ln x+1)^{2}+x^{x-1}.$
(b) $P_{2}(x)=1+(x-1)+(x-1)^{2}.$
(c) $1.1^{1.1}\approx1.11$ and $0.9^{0.9}\approx0.91.$
How did it go?
Stuck? Read 6. Taylor polynomials →
As on the exam
4 problems · 33 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.
An implicit function: its value, its tangent, and which way it goes
The equation $$ y^{3}+xy+x^{2}=1 $$ defines $y$ as a differentiable function of $x$ near the point $x=2{:}$ $y=y(x).$
- (a)3 p
Find all possible real values of $y(2).$
- (b)3 p
Find the equation of the tangent line to the curve at the point $\bigl(2,y(2)\bigr).$
- (c)4 p
Is $y(x)$ increasing or decreasing near $x=2?$ Justify your answer. Then use the tangent line to estimate $y(2.1).$
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(a) A cubic in $y.$ With $x=2$ the equation reads $y^{3}+2y+4=1,$ that is $$ y^{3}+2y+3=0\,. $$
(a) An integer root. An integer root must divide $3$ (Method 1.19), so the candidates are $\pm1$ and $\pm3.$ $y=-1$ gives $-1-2+3=0,$ so $y+1$ is a factor, and division gives $$ \begin{aligned} &y^{3}+2y+3\\ &\quad=(y+1)\bigl(y^{2}-y+3\bigr)\,. \end{aligned} $$
(a) No other real roots. The quadratic factor has discriminant $(-1)^{2}-4\cdot3=-11<0,$ so it has no real roots. Hence $y(2)=-1$ is the only possible value.
(b) Differentiate implicitly. Every $y$ is a function of $x;$ the term $xy$ needs the product rule: $$ 3y^{2}y'+y+xy'+2x=0\,. \tag{1} $$
(b) Insert the point. At $x=2,$ $y=-1{:}$ $$ \begin{aligned} 3y'-1+2y'+4&=0\,,\\ 5y'&=-3\,, \end{aligned} $$ so $y'(2)=-\tfrac35.$
(b) The tangent line through $(2,-1)$ with slope $-\tfrac35{:}$ $$ \begin{aligned} y&=-1-\tfrac35(x-2)\\ &=-\tfrac35x+\tfrac15\,. \end{aligned} $$
(c) The sign of $y'.$ $y'(2)=-\tfrac35<0.$ By (1), $$ y'=-\frac{y+2x}{3y^{2}+x}\,. $$ For $x$ near $2$ the value $y(x)$ is near $-1$ ($y$ is differentiable), so the numerator stays near $3$ and the denominator near $5{:}$ $y'$ stays near $-\tfrac35$ and is negative on an interval around $x=2.$
(c) The verdict. By Fact 2.23, $y$ is (strictly) decreasing near $x=2.$
(c) The estimate. The tangent line is the linear approximation of Fact 4.15 with centre $a=2,$ used at $x=2.1{:}$ $$ \begin{aligned} y(2.1)&\approx-1-\tfrac35\cdot0.1\\ &=-1.06\,. \end{aligned} $$ (Solving $y^{3}+2.1y+3.41=0$ numerically gives $y(2.1)=-1.0587\ldots,$ so the estimate is good to two decimals.)
Answer
(a) $y(2)=-1,$ the only real root of $y^{3}+2y+3=0.$
(b) $y=-1-\tfrac35(x-2),$ that is $y=-\tfrac35x+\tfrac15.$
(c) Decreasing near $x=2,$ since $y'(2)=-\tfrac35<0.$ The estimate is $y(2.1)\approx-1.06.$
Mark your solution
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
- Stopping at the root $y=-1$ without showing that the quadratic factor has no real roots. The problem asks for all possible values, and the solutions of the HT26 exams count the argument as part of the answer.
- Differentiating $xy$ as $xy'$ (or as $y$): the product rule gives $y+xy'.$ Or dropping the $y'$ altogether, as in Watch out 4.10: in the practice problem on the circle fewer than half of the answers were right.
- Losing the sign or turning the fraction over when solving for $y',$ giving $\tfrac35$ or $-\tfrac53.$ In the live vote on the circle these two errors took about half of the answers. Insert your $y'$ into the differentiated equation as a check.
- Reading “decreasing” from $y(2)=-1<0.$ Which way the function goes is the sign of $y',$ not of $y.$
An implicit function: Taylor polynomial of degree 2, and its shape
The equation $$ y^{3}+xy^{2}+x=3 $$ defines $y$ as a twice differentiable function of $x$ near the point $x=1{:}$ $y=y(x).$
- (a)3 p
Find all possible real values of $y(1).$
- (b)5 p
Find the Taylor polynomial of degree $2$ of $y(x)$ about the point $x=1.$
- (c)2 p
Is $y(x)$ increasing or decreasing near $x=1?$ Does its graph bend upwards or downwards there? Justify your answers.
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(a) A cubic in $y.$ With $x=1{:}$ $y^{3}+y^{2}+1=3,$ that is $$ y^{3}+y^{2}-2=0\,. $$
(a) An integer root. The candidates are the divisors of $2$ (Method 1.19). $y=1$ gives $1+1-2=0,$ and division gives $$ \begin{aligned} &y^{3}+y^{2}-2\\ &\quad=(y-1)\bigl(y^{2}+2y+2\bigr)\,. \end{aligned} $$
(a) No other real roots. The discriminant of $y^{2}+2y+2$ is $2^{2}-4\cdot2=-4<0.$ So $y(1)=1$ is the only possible value.
(b) Differentiate once. $xy^{2}$ is a product, and $y^{2}$ gives $2y\,y'{:}$ $$ 3y^{2}y'+y^{2}+2xy\,y'+1=0\,. \tag{1} $$ At $x=1,$ $y=1{:}$ $3y'+1+2y'+1=0,$ so $y'(1)=-\tfrac25.$
(b) Differentiate (1) once more. Every term with $y$ or $y'$ in it is a product of functions of $x{:}$ $$ \begin{aligned} \bigl(3y^{2}y'\bigr)'&=6y(y')^{2}+3y^{2}y''\,,\\ \bigl(y^{2}\bigr)'&=2y\,y'\,,\\ \bigl(2xy\,y'\bigr)'&=2y\,y'+2x(y')^{2}\\ &\quad+2xy\,y''\,, \end{aligned} $$ and the derivative of $1$ is $0.$ The sum is $0.$ $\quad(2)$
(b) Insert $x=1,$ $y=1,$ $y'=-\tfrac25.$ Then $(y')^{2}=\tfrac{4}{25},$ and (2) becomes $$ \begin{aligned} \tfrac{24}{25}+3y''-\tfrac45-\tfrac45&\\ +\tfrac{8}{25}+2y''&=0\,, \end{aligned} $$ so $5y''=\tfrac{8}{25}$ and $y''(1)=\tfrac{8}{125}.$
(b) The Taylor polynomial. By Definition 4.19 with $n=2$ and $a=1,$ the second derivative divided by $2{:}$ $$ \begin{aligned} P_{2}(x)&=1-\tfrac25(x-1)\\ &\quad+\tfrac{4}{125}(x-1)^{2}\,. \end{aligned} $$
(c) Increasing or decreasing. $y'(1)=-\tfrac25<0.$ By (1), $y'=-\dfrac{y^{2}+1}{3y^{2}+2xy},$ which stays near $-\tfrac25$ for $x$ near $1,$ where $y$ stays near $1.$ So $y'<0$ on an interval around $x=1,$ and $y$ is decreasing there (Fact 2.23).
(c) The bending. $y''(1)=\tfrac{8}{125}>0.$ Solving (2) for $y''$ gives a formula in $x,$ $y$ and $y',$ which stays near $\tfrac{8}{125}$ near $x=1.$ So $y''>0$ there, and by Fact 3.22 the graph bends upwards near $x=1.$
Answer
(a) $y(1)=1,$ the only real root of $y^{3}+y^{2}-2=0.$
(b) $y'(1)=-\tfrac25$ and $y''(1)=\tfrac{8}{125},$ so $P_{2}(x)=1-\tfrac25(x-1)+\tfrac{4}{125}(x-1)^{2}.$
(c) Decreasing near $x=1$ ($y'<0$), and the graph bends upwards ($y''>0$).
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 10 p
Common errors
- In the second differentiation, treating $y'$ as a constant: the term $3y^{2}y'$ then gives only $6y(y')^{2}$ and no $y''$ at all, and the equation cannot be solved for $y''.$ A $y'$ leaves a $y''$ behind, as a $y$ leaves a $y'.$
- Losing one of the three terms of $\bigl(2xy\,y'\bigr)',$ a product of three functions of $x.$ Write it as $(2x)\cdot(y\,y')$ and use the product rule twice.
- Forgetting to divide $y''(1)$ by $2$ in the Taylor polynomial, or writing powers of $x$ instead of $(x-1).$ In the practice problem on the Taylor polynomial of degree $2$ only about half of the answers were right.
- Inserting the numbers into (1) before differentiating it a second time. After $x=1,$ $y=1$ are inserted, (1) is no longer an equation that holds for every $x,$ and its derivative means nothing.
An implicit function with an exponential, in one line
The equation $$ e^{y}+(x+1)\,y=x^{2} $$ defines $y$ as a twice differentiable function of $x$ near $x=1{:}$ $y=y(x).$
- (a)7 p
Find the Taylor polynomial of degree $2$ of $y(x)$ about $x=1.$
- (b)1 p
Does the graph of $y(x)$ bend upwards or downwards near $x=1?$
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(a) The value, by inspection. At $x=1$ the equation reads $e^{y}+2y=1.$ Clearly $y=0$ is a solution, since $e^{0}+0=1.$
(a) Differentiate once. $e^{y}$ gives $e^{y}y'$ by the chain rule, and $(x+1)y$ is a product: $$ e^{y}y'+y+(x+1)\,y'=2x\,. \tag{1} $$ At $x=1,$ $y=0{:}$ $y'+0+2y'=2,$ so $y'(1)=\tfrac23.$
(a) Differentiate (1) once more. $e^{y}y'$ is a product, and so is $(x+1)y'{:}$ $$ \begin{aligned} e^{y}(y')^{2}+e^{y}y''+y'&\\ +\,y'+(x+1)\,y''&=2\,. \end{aligned} $$
(a) Insert $x=1,$ $y=0,$ $y'=\tfrac23.$ $$ \tfrac49+y''+\tfrac43+2y''=2\,, $$ so $3y''=2-\tfrac{16}{9}=\tfrac29$ and $y''(1)=\tfrac{2}{27}.$
(a) The Taylor polynomial. By Definition 4.19, $$ \begin{aligned} P_{2}(x)&=0+\tfrac23(x-1)\\ &\quad+\tfrac{1}{27}(x-1)^{2}\,. \end{aligned} $$
(b) The bending. $y''(1)=\tfrac{2}{27}>0,$ and the formula for $y''$ from the last equation, built from $x,$ $y$ and $y',$ keeps that sign near $x=1.$ By Fact 3.22 the graph bends upwards near $x=1.$
Looking ahead — not in the lecture notes at this point. In the words of Lecture 6, $y$ is convex near $x=1;$ that is how an exam asks it.
Answer
(a) $y(1)=0,$ $y'(1)=\tfrac23,$ $y''(1)=\tfrac{2}{27},$ so $P_{2}(x)=\tfrac23(x-1)+\tfrac{1}{27}(x-1)^{2}.$
(b) It bends upwards, since $y''(1)>0.$
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Common errors
- Writing the derivative of $e^{y}$ as $e^{y}.$ It is $e^{y}y'{:}$ the same error as Watch out 4.10, with $e^{y}$ in place of $y^{2}.$
- Giving $y(1)=0$ with no argument that it is the only solution. With an exponential there is no discriminant to compute; the argument is that $e^{y}+2y$ is strictly increasing.
- Differentiating $e^{y}y'$ as $e^{y}y''$ only, which loses $e^{y}(y')^{2}$ and gives $y''(1)=\tfrac{2}{9}$ instead of $\tfrac{2}{27}.$
- Writing $\tfrac{2}{27}$ as the coefficient of $(x-1)^{2}{:}$ the second derivative is divided by $2!=2.$
Taylor polynomials of a logarithm and of a root, and an estimate
Let $f(x)=\ln(1+2x)$ for $x>-\tfrac12,$ and $g(x)=\sqrt{4+x}$ for $x>-4.$
- (a)2 p
Find the Taylor polynomial of degree $3$ of $f$ about $x=0.$
- (b)3 p
Find the Taylor polynomial of degree $2$ of $g$ about $x=0,$ and use it to estimate $\sqrt{4.2}.$
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(a) Three derivatives. By (4.2), and then the power rule with the chain rule, $$ \begin{aligned} f'(x)&=\frac{2}{1+2x}\,,\\ f''(x)&=-\frac{4}{(1+2x)^{2}}\,,\\ f'''(x)&=\frac{16}{(1+2x)^{3}}\,. \end{aligned} $$ At $0{:}$ $f(0)=0,$ $f'(0)=2,$ $f''(0)=-4$ and $f'''(0)=16.$
(a) The polynomial. Divide by $1!,$ $2!=2$ and $3!=6{:}$ $$ P_{3}(x)=2x-2x^{2}+\tfrac83x^{3}\,. $$
(b) Two derivatives. $g(x)=(4+x)^{1/2},$ so by Fact 4.14 $$ \begin{aligned} g'(x)&=\tfrac12(4+x)^{-1/2}\,,\\ g''(x)&=-\tfrac14(4+x)^{-3/2}\,. \end{aligned} $$ At $0{:}$ $g(0)=2,$ $g'(0)=\tfrac12\cdot\tfrac12=\tfrac14$ and $g''(0)=-\tfrac14\cdot\tfrac18=-\tfrac{1}{32}.$
(b) The polynomial. $g''(0)$ divided by $2{:}$ $$ Q_{2}(x)=2+\tfrac14x-\tfrac{1}{64}x^{2}\,. $$
(b) The estimate. $\sqrt{4.2}=g(0.2),$ so the centre is $0$ and $x=0.2{:}$ $$ \begin{aligned} \sqrt{4.2}&\approx2+0.05-0.000625\\ &=2.049375\,. \end{aligned} $$ (The true value is $2.04939\ldots$)
Answer
(a) $P_{3}(x)=2x-2x^{2}+\tfrac83x^{3}.$
(b) $Q_{2}(x)=2+\tfrac14x-\tfrac{1}{64}x^{2},$ and $\sqrt{4.2}\approx2.049375.$
Mark your solution
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
- Losing the inner derivative $2$ of $1+2x{:}$ $f'(x)=\dfrac{1}{1+2x}$ instead of $\dfrac{2}{1+2x}.$ Then every coefficient of $P_{3}$ is wrong. The factor $2$ comes back at each differentiation, which is why $f'''(0)=16$ and not $2.$
- Inserting $x=4.2$ instead of $x=0.2.$ The centre is $0,$ where $g(0)=\sqrt4$ is known, and the polynomial is evaluated at the distance from it: the $a$ and $x$ of Fact 4.15 mixed up.
- Adding $1$ to the exponent instead of subtracting $1$: $(4+x)^{-1/2}$ differentiated to $(4+x)^{1/2}$ times something. The exponent goes $-\tfrac12\mapsto-\tfrac32.$