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Math 105: Review for Exam II - Solutions
- Find dy/dxdy/dxdy/dx for each of the following.
(a) y = x^2 + 2x^ + e^2 + e^2 x^ + ln 2 + ln (2x) + arctan 2 dy dx
= 2x + (ln 2)2x^ + 2e^2 x^ +
2 x
· 2 Note that e^2 , ln 2, and arctan 2 are constants.
(b) y =
x · arctan (5x) dy dx
x−^1 /^2 arctan(5x) +
x ·
1 + (5x)^2
arctan(5x) 2 x^1 /^2
x 1 + 25x^2
(c) y = ln(tan(2cos(x
(^2) ) )) dy dx
tan(2cos(x^2 ))
· sec^2 (2cos(x
(^2) ) ) · ln 2(2cos(x
(^2) ) ) · (− sin(x^2 )) · 2 x
(d) y = sin^5
x + eπ ln 4 + arcsin 6x
Note that sin^5 w = (sin w)^5.
dy dx = 5 sin^4
x + eπ ln 4 + arcsin 6x
·cos
x + eπ ln 4 + arcsin 6x
(1)(ln 4 + arcsin 6x) − (x + eπ^ )( √^1 1 −(6x)^2
(ln 4 + arcsin 6x)^2
(e) y = (x^2 + 1)sin^ x^ We need logarithmic differentiation here. ln y = sin x · ln(x^2 + 1) Take ln of each side. 1 y
dy dx = cos x · ln(x^2 + 1) + sin x ·
x^2 + 1 · 2 x Differentiate.
dy dx
[
cos x · ln(x^2 + 1) +
2 x sin x x^2 + 1
]
y Solve for
dy dx
dy dx
[
cos x · ln(x^2 + 1) +
2 x sin x x^2 + 1
]
· (x^2 + 1)sin^ x^ Replace y.
- Consider the curve defined by x^3 + y^3 =
x^3 + y^3 = xy
x^3 + y^3 = xy
xy (known as the Folium of Descartes).
(a) Find dy/dxdy/dxdy/dx. Use implicit differentiation.
3 x^2 + 3y^2 dy dx
y +
x dy dx 3 y^2
dy dx
x
dy dx
y − 3 x^2 dy dx
3 y^2 −
x
y − 3 x^2
dy dx
9 2 y^ −^3 x
2
3 y^2 − 92 x
(b) Find the equation of the tangent line at the point (1,2). We want y = mx + b.
m =
9 2 ·^2 −^3 ·^1
2 3 · 22 − 92 · 1
, so y =
x + b.
Now plug in x = 1 and y = 2 to find b. 2 =
· 1 + b ⇒
= b
Therefore, we have y =
x +
- Evaluate the following limits.
Throughout this solution, the symbol F will stand for whatever notation your instructor prefers for using L’Hopital’s Rule on the indeterminate form 0/0; this may be “0/0” = or L
′H = or H = or = “0/0” or “has the form ‘
’ and so, by L’Hopital’s Rule, is equal to” or something else. The symbol ♥ will serve the same purpose for the indeterminate form ∞/∞.
(a) lim x→ 0
sin 3x 5 x
lim x→ 0
sin 3x 5 x lim x→ 0
sin 3x 5 x F lim x→ 0
3 cos 3x 5
(b) lim x→∞
ex ln x
lim x→∞
ex ln x
lim x→∞
ex ln x
♥ lim x→∞
ex 1 /x
= lim x→∞ xex^ = ∞
(c) lim x→ 0
1 − cos 2x 3 x
lim x→ 0
1 − cos 2x 3 x lim x→ 0
1 − cos 2x 3 x^
= 0 Can’t use (and don’t need) L’Hopital’s Rule!
(d) lim x→ 1
x^3 − 1 7 − 7 x
lim x→ 1
x^3 − 1 7 − 7 x lim x→ 1
x^3 − 1 7 − 7 x
F lim x→ 1
3 x^2 − 7
(e) lim x→ 0
1 − cos 4x 5 x^2
lim x→ 0
1 − cos 4x 5 x^2 lim x→ 0
1 − cos 4x 5 x^2 F lim x→ 0
4 sin 4x 10 x F lim x→ 0
16 cos 4x 10
- Suppose the domain of fff(((xxx))) is all reals and that fff has an inverse function fff−−−^111 (((xxx))). Further, suppose that fff(2) = 5(2) = 5(2) = 5 and fff′′′(2) =(2) =(2) = eee. Finally, let hhh(((xxx) = 1) = 1) = 1/f/f/f(((xxx))).
(a) What point must be on the graph of fff−−−^111 (((xxx)))? Since f(2) = 5, we also know that f−^1 (5) = 2; therefore, the point (5,2) is on the graph of f−^1 (x). (b) What point must be on the graph of hhh(((xxx)))? Since f(2) = 5, we also know that h(2) = 1/5; therefore, the point (2,1/5) is on the graph of h(x). The moral of the story in parts (a) and (b) is that inverses and reciprocals are not the same. (c) Give an example of a point that cannot be on the graph of fff(((xxx))). Do not choose a point with xxx-value of 2. Since f−^1 (x) is a function, we know that for each of its inputs (such as 5) there is only one output (in this case, 2). This means that no point of the form (5, k) can be on the graph of f−^1 (x) (except for k = 2). Therefore, no point of the form (k, 5) can be on the graph of f(x) (except for k = 2). [Note: this is the same as saying that f must be a one-to-one function; on f, no two x-values correspond to the same y-value.] (d) What is the value of the derivative of hhh(((xxx))) at xxx = 2= 2= 2?
By the Quotient Rule (or Chain Rule if you prefer), h′(x) =
0 · f(x) − 1 · f′(x) [f(x)]^2
f′(x) [f(x)]^2
So, h′(2) = −
f′(2) [f(2)]^2
e 52
e 25
(e) What is the value of the derivative of fff−−−^111 (((xxx))) at xxx = 5= 5= 5? Since f and f−^1 are inverses of one another, we know that f(f−^1 (x)) = x.
f(f−^1 (x)) = x f′^ (f−^1 (x)) · [f−^1 (x)]′^ = 1 Differentiate both sides. Use the Chain Rule on the left.
[f−^1 (x)]′^ =
f′(f−^1 (x))
Now solve for [f−^1 (x)]′.
Using this final formula, we see that the derivative of f−^1 (x) at x = 5 is
f′(f−^1 (5))
f′(2)
e
(d) Find the xxx-value(s) of all inflection points.
f′′(x) = 12x^2 ex^ + 4x^3 ex^ + 4x^3 ex^ + x^4 ex^ Use Product Rule on each product in f′(x) above. 0 = ex^ (x^4 + 8x^3 + 12x^2 ) 0 = ex^ x^2 (x^2 + 8x + 12) 0 = ex^ x^2 (x + 2)(x + 6) ⇒ x = 0, − 2 , − 6
x < − 6 − 6 < x < − 2 − 2 < x < 0 0 < x f′′^ positive negative positive positive f concave up concave down concave up concave up So, the x-values of the inflection points of f are x = −2 and x = −6 but NOT x = 0.
(e) Sketch fff.
0 -12 -6 -4 -2 0
4
10
8 6
2
- How would your answers to the previous question change if the domain of fff were [[[−−− 101010 ,,, 10]10]10]?
There would be a global maximum at (10, 104 e^4 ). (And the graph would be restricted to − 10 ≤ x ≤ 10).
- Circle always, sometimes, or never to make each statement below correct.
(a) If fff′′′(1) = 0(1) = 0(1) = 0 then fff always /sometimes/never has a critical point at xxx = 1= 1= 1. A critical point is where f′^ is 0 or undefined.
(b) If fff′′′(2) = 0(2) = 0(2) = 0 then fff always/ sometimes /never has a local maximum or local minimum at xxx = 2= 2= 2. f might instead have a terrace point at x = 2; we need f′^ to change sign at x = 2 in order to guarantee a local extremum there.
(c) If xxx = 3= 3= 3 is a critical point of fff , then fff′′′^ (3)(3)^ (3) is always/ sometimes /never 0. It may also be that f′(3) is undefined.
(d) If fff′′′′′′(4) = 0(4) = 0(4) = 0, then fff always/ sometimes /never has an inflection point at xxx = 4= 4= 4. We need f′′^ to change sign at x = 4 to guarantee an inflection point there. For example, if f(x) = (x−4)^4 then f′′(4) = 0 but f has a local minimum rather than an inflection point at x = 4. However, if f(x) = (x − 4)^3 then f′′(4) = 0 and f does have an inflection point at x = 4. Also see what happens at x = 0 in problem 7(d).
(e) If fff has a global maximum at xxx = 5= 5= 5, then fff′′′(5)(5)^ (5) is always/ sometimes /never 0. f′^ (5) might also be undefined, or x = 5 might be an endpoint of the domain.
(f) If fff′′′(6) = 0(6) = 0(6) = 0 and fff′′′′′′(6) =(6) =(6) = −−− 222 , then fff always /sometimes/never has a local maximum at xxx = 6= 6= 6. If f is concave down with a horizontal slope at x = 6, then f must have a local maximum there.
(g) If fff′′′^ (7) = 0(7) = 0^ (7) = 0 and fff′′′′′′(7) = 0(7) = 0(7) = 0, then fff always/ sometimes /never has a local extremum at xxx = 7= 7= 7. This means that the second derivative test is inconclusive, so you need to use a different test. For example, if f(x) = (x − 7)^4 then f′^ (7) = 0 and f′′(7) = 0 and f has a local minimum at x = 7. However, if f(x) = (x − 7)^3 then f′(7) = 0 and f′′(7) = 0 and f has an inflection point but not a local extremum at x = 7.
- The rate of change of a population PPP (((ttt))) of eels is proportional to the size of the population. When the population is 40000, it is growing at a rate of 400 eels per year. At time ttt = 0= 0= 0, the population is 10000.
(a) Write a differential equation whose solution is PPP (((ttt))). Rate of change (P ′) is (=) proportional to (k) size of population (P ) means P ′^ = kP. What’s the value of k? When P = 40000, we know P ′^ = 400. That is, 400 = k · 40000, so k = .01. Thus, we have P ′^ =. 01 P. (b) Solve your differential equation. The general solution is P (t) = Ae.^01 t. What’s the value of A? When t = 0, we know P = 10000. That is, 10000 = Ae^0 = A, so A = 10000. Thus, we have P (t) = 10000e.^01 t. (c) When will the population reach 60000?
60000 = 10000e.^01 t 6 = e.^01 t ln 6 = ln e.^01 t^ Take ln of each side. ln 6 =. 01 t Recall that ln ez^ = z. t = 100 ln 6 ≈ 179 .176 years
- Use Newton’s Method with an initial guess of xxx 000 === −−− 111 to find the next three approxima- tions to a solution of xxx^333 +++ xxx + 1 = 0+ 1 = 0+ 1 = 0. Then test your final approximation to see if it appears to be close to a root.
Recall that xn+1 = xn −
f(xn ) f′(xn)
= xn −
x^3 n + xn + 1 3 x^2 n + 1
Using a calculator with initial guess x 0 = −1, we get the following.
x 0 = − 1 x 1 = − 0. 75 x 2 = − 0. 68604 ... x 3 = − 0. 68233 ...
Check: (− 0 .68233)^3 + (− 0 .68233) + 1 ≈ 0 .0000526, so our approximation seems to be a good one.
