Dear AP Calculus BC student,
Hello and welcome to the wonderful world of AP Calculus! We are excited that you have
elected to take an accelerated mathematics course such as AP Calculus BC and would like to
welcome you to what will surely be a challenging yet rewarding year of mathematics. AP
Calculus BC is intended for students who have a thorough knowledge of algebra, geometry,
trigonometry and basic functions. The functions you must be familiar with include linear,
polynomial, exponential, logarithmic, trigonometric, and piecewise-defined. In particular, before
studying calculus, you must be familiar with the properties, algebra, and graphs of functions.
Knowledge of the basic trigonometric identities is also essential, along with the values of the
trigonometric functions of common angles.
We would like to help you get a good start in AP Calculus BC. These problems are to be
completed over the summer, and it is our expectation that all problems will be thoughtfully
attempted, with all work and justification shown. Additionally, there is a significant amount of
memorization expected for success on the AP exams. It is expected that students will return to
school having mastered those procedural basics to permit more time to deeply understand the
critically important conceptual aspects of the course. Also attached are 6 mostly new and skillbased calculus topics students are expected to master prior to the start of the school year.
Please plan to devote approximately 2 hours of work to the review questions and at least 1 hour
of work to each of the 6 new content topics.
We look forward to supporting you in a successful and rewarding year in mathematics!
Sincerely,
The AP Calculus BC teachers
Anticipated work load: 8 hours
Summer Packets are due the FIRST FRIDAY of school 2014-2015
Summer Assignment Quiz the same day
Serving the communities of Antioch, Lake Villa, Lindenhurst, and Old Mill Creek
Community High School District 117, being a community of learners with a vision of excellence, is committed to providing an educational experience
that encourages all learners to develop to their fullest potential, to engage in lifelong learning, and to be responsible members of society.
Algebra Key Concepts 1. Write the equation of the line that passes through the points ( 3 , 1 ) and ( -­‐5, -­‐3 ) in point-­‐slope form then convert it to slope-­‐intercept form. 2. Write an equation of the line that is the perpendicular bisector of the segment connecting an x-­‐intercept of 2 and a y-­‐intercept of 6 then convert it to slope-­‐intercept form. 3. Find the points of intersection of the line y = x + 1 and the parabola y = x2 – 11 using algebraic methods. Show work. !
4. If 𝑓 𝑥 = and 𝑔 𝑥 = 𝑥 ! + 2𝑥 find 𝑓 𝑔 𝑥 and 𝑔 𝑓 𝑥 . !
Geometry Key Concepts 5. If the area of a rectangle is 𝑥 ! − 3𝑥 − 4 and the height of the rectangle is 𝑥 + 1, find an expression for the length of the base. !
6. If the volume of a sphere is 36π 𝑉 = 𝜋𝑟 ! , find the surface area 𝑆𝐴 = 4𝜋𝑟 ! . !
!
M#!!!C&.8!')(!=-0+;(!-/!,!<-.(!4! ! !! ! !7!&/!')(!%,'&-!-/!')(!)(&3)'!'-!')(!%,8&+2!&2!5N"!,.8!')(!<-.(!),2!,!
!
)(&3)'!-/!">!<;#!
O#!!!C&.8!')(!,%(,!-/!')(!0,%3(%!'%,1(P-&8!&/!
')(!2),1(2!,%(!2&;&0,%#!
:!
J!
M!
Q#!!!C&.8!')(!2),8(8!,%(,!&/!')(!<&%<0(2!,%(!<-.<(.'%&<6!!
,.8!')(!<&%<+;/(%(.<(!-/!')(!0,%3(%!<&%<0(!&2!:OL6!,.8!!
RS6!H)&<)!&2!',.3(.'!'-!')(!2;,00(%!<&%<0(6!&2!J>!+.&'2#!
B!
S!
R!
W!
"45(-.%4!"#$%&'(!)%*!+,-.%/01!
6,#5%!7,'!(##!5(#8%1!,7!9:!;<%'%!9!=1!(!'%(#!-83&%'>!!!?,!-,0!81%!(!.(#.8#(0,'!8-#%11!4='%.0%4>!!@%(5%!(-1;%'1!
%9(.0>!
!
"T#!!!! ! !! ! ! !!
""#!!!!"!! ! ! !!! !
">#!!!!"# ! ! ! ! ! !!
"5#!!!! ! ! !"# !!!!
UV2(!,!<,0<+0,'-%#!
Simplify without a calculator: !
14. ln 𝑒 ! !
!
15. 5𝑎 !
4𝑎 ! ) 17. sec arcsin
19. csc
16. sin!! (−
!
!
!
!"
!!
18. cot
!
!!
!
Sketch a quick graph of the function labeling key values (calculator allowed) and determine whether the function is even or odd (explain what feature on the graph indicates even v. odd): 20. 𝑔 𝑥 = −3 cos
!
!
21. 𝑓 𝑥 = 𝑥 ! − 𝑥 22. 𝑓 𝑥 =
2 − 𝑥,
𝑥 !,
𝑥≤1
𝑥>1
a. Graph the function. Label the coordinates of at least three points on each graph. b. Domain: c. Range: d. 𝑓 2 = e. 𝑓 𝑓 −7 = Graph each function. Label any vertical and horizontal asymptotes, holes, x and y-­‐intercepts: !!!!
24. ℎ 𝑥 = !
25. 𝑘 𝑥 = ln(𝑥 + 2) ! !!!!
–11–10 –9 –8
Domain: Range: 11
y
11
10
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
–7 –6 –5 –4 –3 –2 –1
–1
1
2
3
4
5
6
7
8
9 10
x
–11–10 –9 –8 –7 –6 –5 –4 –3 –2 –1
–1
–2
–2
–3
–3
–4
–4
–5
–5
–6
–6
–7
–7
–8
–8
–9
–9
–10
–10
–11
–11
Domain: Range:
y
1
2
3
4
5
6
7
8
9 10
x
DEF!GH@@HIJK2!L"2F6!+HKD"JK!
KFI!+HKDFKD!DH!M"6DFN!OFGHNF!6+EHH@!6D"ND6!
! !!
\]^!\_`KWG!_a!\]^!C_YY_$KaX!`RX^G!R[^!b_G\Yc!a^$#!!!
]_a^G\!W_YYRS_[R\K_a!$K\]!C[K^aZG!KG!^aW_V[RX^Z#!!
`Y^RG^!W_aGVY\!\]^!dKZ^_!YKae!`[K_[!\_!RGeKaX!C_[!]^Y`#!
\]R\!GRKZ6!`Y^RG^!RGe!C_[!]^Y`#!
b[#!Y^d^[^a\f!$KYY!S^!g!YW]G!ZV[KaX!\]^!GVbb^[#!
CKaZ!R!WRYWVYVG!C[K^aZ!\_!S[KaX!RaZ!^bRKY!b[#!Y^d^[^a\f!$K\]!$]^a!c_VhZ!YKe^!\_!X^\!\_X^\]^[!\_!
G\R[\!$_[eKaX!_a!WRYWVYVG#!!!
GW_\\#Y^d^[^a\fgW]GZ""M#_[X!
Z_ah\!`[_W[RG\KaR\^i!
S^WRVG^!\]^!WV[[KWVYVb!_C!R`!W_V[G^G!KG!CRK[Yc!VaKd^[GRY6!^jW^YY^a\![^G_V[W^G!RaZ!dKZ^_G!R[^!
RdRKYRSY^!_aYKa^#!
KhZ!GVXX^G\!G\R[\KaX!]^[^N!)''1Nkk1,'%&<Il;'#<-;km<,0<+0+2!
^d^[c!GKaXY^!WRYWVYVG!G\VZ^a\!$KYY!a^^Z!]^Y`!R\!G_b^!`_Ka\!\]KG!c^R[!F!Y^\G!X^\!_CC!_a!\]^![KX]\!
C__\!RaZ!RGe!C_[!]^Y`!^R[Yc!RaZ!_C\^ai!
GKXa!V`!C_[![^bKaZ"T"!Raa_VaW^b^a\G!_C!GVbb^[!]^Y`!G^GGK_aG!
!
Concept #1 – Limits Learn – 1. How to evaluate one-­‐sided limits 2. How to evaluate two-­‐sided limits 3. How to evaluate limits at infinity 4. When a limit equals infinity v. negative infinity v. does not exist (hint: not the same thing) Watch – https://www.youtube.com/watch?v=zpcBkRpqHqQ, http://patrickjmt.com/topic/limits/page/2/ Try – Limit hints – try the following approaches in order: 1. “Plug it in, plug it in” – substitute the x value into the function. If the solution is a real number, ! ∞
congrats, you’re done. If it is , , 0 ∗ ∞ and other confusing forms called indeterminate ! ∞
(http://en.wikipedia.org/wiki/Indeterminate_form). It does NOT equal 0! It means the value cannot be determined without further algebraic analysis. 2. If substituting the x-­‐value yields an indeterminate solution, try one of the algebraic strategies below depending on the function form: a. Rational functions – factor the numerator and denominator, simplify, and substitute again. b. Compound rational expresions – work towards common denominators for fractions being added or subtracted then refer to (a). c. Radical expressions – multiply by the conjugate of the radical expression (i.e. 𝑎 + 𝑏 ℎ𝑎𝑠 𝑎 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒 𝑜𝑓 ( 𝑎 − 𝑏)). 3. Analyze the limit graphically. !"# !
0. Sketch a graph of 𝑓 𝑥 =
. !
a. 𝐹𝑖𝑛𝑑 lim!→! 𝑓(𝑥) from the graph. b. Find lim!→∞ 𝑓(𝑥) from the graph. c. Memorize either the two limits above or the graph of 𝑓 𝑥 above. d. lim!→!
!"# !!
!!
(Graph and make logical conclusion about future
!"# !"
!"
limits. Evaluate each of the following limits without a graphing calculator using the hints above:
1. lim!→!! 𝑥 ! − 2𝑥 ! + 1
!
2. lim!→!!
!
!!
3. lim!→! !!!!
! ! !! !
5. lim!→! !"! ! !!"#
4. lim!→!
! ! !!
!! ! !!!!!
!! ! !!
!! ! !!
6. lim!→!∞
!! ! !!!!!
! ! !!!!!
7. lim!→!!
9. lim!→!
!!!"# !
!!!"# !
!!!!!
!!!
Use a graph:
!
11. lim!→!
!
8. lim!→!
!! ! !!!!!
! ! !!!!!
10. lim!→!
!"!!
!!! !!
12. lim!→! −
+4
!
!!
13. Given the function f x whose graph is shown below (ticks marks in picture have scale 1): a) lim!→!! 𝑓 𝑥 = e) 𝑓 3 = b) lim!→!! 𝑓 𝑥 = f) 𝑓 −1 = d) lim!→!!! 𝑓 𝑥 = g) lim!→!! 𝑓 𝑥 = e) lim!→!!! 𝑓 𝑥 = h) lim!→! 𝑓 𝑥 = 14. Sketch a graph of a function ℎ(𝑥) that has the following properties: a. lim!→!! ℎ 𝑥 = 0 b. lim!→ !!! ℎ 𝑥 = −∞ c. lim!→ !!! ℎ 𝑥 = ∞ d. lim!→! ℎ 𝑥 = 0 e. lim!→! ℎ 𝑥 = 1 (4 pts) 15. Sketch a possible graph for a function f that has the stated properties. f(-2) exists, the limit as x
approaches -2 exists, f is not continuous at x = -2, and the limit as x approaches 1 does not exist.
Concept #2 – Definition of Derivative Learn – 1. What a definition of derivative represents with respect to rate of change and the function 2. How to apply the standard definition of derivative lim!→!
lim!→!
! !!! !! !!!
!!! ! !!!
! !!! !! !
!!! !!
or other definitions , 𝑒𝑡𝑐. 3. How a derivative compares to an average rate of change Watch – http://patrickjmt.com/understanding-­‐the-­‐definition-­‐of-­‐the-­‐derivative/, http://patrickjmt.com/the-­‐
difference-­‐quotient-­‐example-­‐2/ Try – Use the definition of derivative 𝑓 ! 𝑥 = lim!→!
! !!! !! !
!!! !!
to determine 𝑓′(𝑥) for each of the functions below. Use 𝑓′(𝑥) to find the slope of the function at the x-­‐value given, then write a tangent line in point-­‐slope form. 1. 𝑓 𝑥 = 3𝑥 − 1 (hint: if you understand what a derivative is, you don’t even need to use the limit for part a) a. Find 𝑓′(𝑥). b. Find 𝑓′(3). c. Find the equation of a tangent line at x = 3. d. Sketch 𝑓 𝑥 and the tangent line at x = 3. 2. 𝑓 𝑥 = 𝑥 ! a. Find 𝑓′(𝑥). b. Find 𝑓′(−2). c. Find the equation of a tangent line at x = -­‐2. d. Sketch 𝑓 𝑥 and the tangent line at x = -­‐2. 3. 𝑓 𝑥 = 2𝑥 ! − 5𝑥 a. Find 𝑓′(𝑥). b. Find 𝑓′(1). c. Find the equation of a tangent line at x = 1. d. Sketch 𝑓 𝑥 and the tangent line at x = 1. 4. 𝑓 𝑥 = 2𝑥 ! (hint: 𝑎 + 𝑏 ! = 𝑎 ! + 3𝑎 ! 𝑏 + 3𝑎𝑏 ! + 𝑏 ! ) a. Find 𝑓′(𝑥). b. Find 𝑓′(−1). c. Find the equation of a tangent line at x = -­‐1. d. Sketch 𝑓 𝑥 and the tangent line at x = -­‐1. 5. Carefully compare the original functions f(x) in each of the problems above to the derivatives f’(x). There is a fairly simple and incredibly powerful pattern called the Power Rule. Try to describe it in 1-­‐2 sentences below. +,-.%/0!PY!R!L,;%'!N8#%!
"#$%&!'!!
"#!!!]-H!'-!*+&<I0A!8('(%;&.(!')(!8(%&=,'&=(!-/!,!1-0A.-;&,0!/+.<'&-.#!
>#!!!]-H!'-!,110A!')(!`-H(%![+0(!/-%!2&;10(!%,'&-.,0!,.8!%,8&<,0!/+.<'&-.2#!
($)*+!'!+)),-.//0001234)45#1*36/0$)*+789FLMLN<E-4O*A!+)),.//234)415#/(%(P0MEQR=-!
G%2!'!!
1. Once you’ve identified the Power Rule, try to write it in the box below, the find f’(x) for each
function below by applying the Power Rule.
`-H(%![+0(N!!
K/!!!!! ! ! ! 6!')(.!! !!!! !pppppppppppppp!
b. ! ! ! !! ! ! !! ! !
a. ! ! ! !! ! !
c. ! ! ! !! ! ! !! ! !""""""""
d. ! ! ! !! !"
e. ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! ! !
f. ! ! !
g. ! ! !
i. ! ! !
!
(hint: express in the form ! ! first)
!
!
h. ! ! ! !
!!
j. ! ! !
!!
!
!
!
!
!
!
2. ! !! ! represents the “second derivative” of the function f(x) – the derivative of the derivative.
Similarly, !!!!!!! represents the third derivative, and for the fourth derivative and higher (since the
apostrophes would get a bit ridiculous) derivatives are expressed as ! ! !!!, etc.
Find ! !! ! ! ! !!! ! ! !
!
! !!
!
! ! !"#!!
!
!!! for the function in (e) above.
For the three functions below, find the average rate of change (ARoC =
!
! ! !! !
!!!
) and instantaneous
rate of change (𝑓 𝑥 using the Power Rule), then use 𝑓′(𝑥) to write a tangent line and find where the
slope has a particular value.
3. 𝑏 𝑥 = 5𝑥 ! − 2𝑥
Average RoC over [ 0 , 2 ] ?
Instantaneous RoC expression:
X-value where slope is -12?
Tangent line @ x = 0?
4. 𝑒 𝑥 = 𝑥 ! − 5𝑥
Average RoC over [ -3 , 3 ] ?
Instantaneous RoC expression:
X-value where slope is 7?
Tangent line @ x = 1?
!
5. 𝑓 𝑥 = + 3
!
Average RoC over [ 1 , 2 ] ?
Instantaneous RoC expression:
X-value where slope is -1 ?
Tangent line @ x = 4 ?
6. Find the point(s) on the curve 𝑦 = 2𝑥 ! − 3𝑥 ! − 12𝑥 where the tangent line is horizontal. +,-.%/0!PZ!R!+,-0=-8=0*!(-4!?=77%'%-0=(&=#=0*!
"#$%&!'!!
"#!!!]-H!'-!8('(%;&.(!H)(%(!,!/+.<'&-.!&2!<-.'&.+-+2!,.8!8&//(%(.'&,?0(!
>#!!!]-H!'-!2-0=(!/-%!,!;&22&.3!=,0+(!&.!,!1&(<(H&2(!/+.<'&-.!3&=(.!'),'!')(!/+.<'&-.!&2!<-.'&.+-+2!,.8!
8&//(%(.'&,?0(!
5#!!!$),'!'A1(2!-/!8&2<-.'&.+&'&(2!(@&2'!,.8!H),'!<,+2(2!/+.<'&-.2!'-!.-'!?(!8&//(%(.'&,?0(!
($)*+!'!+)),.//234)415#/ST4BU,">V%WA!+)),.//234)415#/MXYZ[8(\8Y]A!+)),.//234)415#/YB^,_+W6"_0!
G%2!'!!
?%7=-=0=,-!,7!+,-0=-8=0*A!!
\)(!/+.<'&-.!/!&2!<-.'&.+-+2!,'!2-;(!1-&.'!<!-/!&'2!8-;,&.!&/!')(!0&;&'!-/!/4@7!,2!@!,11%-,<)(2!<!')%-+3)!')(!
8-;,&.!-/!/!(@&2'2!,.8!&2!(*+,0!'-!/4<7#!!!K.!;,')(;,'&<,0!.-','&-.6!')&2!&2!H%&''(.!,2!!
!!K.!8(',&0!
')&2!;(,.2!')%((!<-.8&'&-.2N!/&%2'6!/!),2!'-!?(!8(/&.(8!,'!<#!G(<-.86!')(!0&;&'!-/!'),'!(*+,'&-.!),2!'-!(@&2'#!\)&%86!
')(!=,0+(!-/!')&2!0&;&'!;+2'!(*+,0!/4<7#!!\)(!/+.<'&-.!/!&2!2,&8!'-!?(!<-.'&.+-+2!&/!&'!&2!<-.'&.+-+2!,'!(=(%A!1-&.'!-/!
&'2!8-;,&.#!!`%&;,%A!4=1.,-0=-8=0=%1!,%(!1-&.'!4)-0(76!l+;16!&./&.&'(6!,.8!-2<&00,'&.3!8&2<-.'&.+&'&(2#!!!
?%7=-=0=,-!,7!?=77%'%-0=(&=#=0*A!
K.!<,0<+0+26!,!8&//(%(.'&,?0(!/+.<'&-.!-/!-.(!%(,0!=,%&,?0(!&2!,!/+.<'&-.!H)-2(!8(%&=,'&=(!(@&2'2!,'!(,<)!1-&.'!&.!
&'2!8-;,&.#!R2!,!%(2+0'6!')(!3%,1)!-/!,!8&//(%(.'&,?0(!/+.<'&-.!;+2'!),=(!,!.-.9=(%'&<,0!',.3(.'!0&.(!,'!(,<)!1-&.'!
&.!&'2!8-;,&.6!?(!%(0,'&=(0A!2;--')6!,.8!<,..-'!<-.',&.!,.A!8&2<-.'&.+&'&(26!<-%.(%26!=(%'&<,0!',.3(.'26!-%!<+212#!
"#!!!C&.8!')(!=,0+(!-/!!!2-!'),'!')(!/+.<'&-.!&2!<-.'&.+-+2#!!4]&.'N!\)(!'H-!1,%'2!.((8!'-!),=(!')(!2,;(!A9=,0+(!,'!
')(!@9=,0+(!H)(%(!')(A!H-+08!n;(('o#7!
! ! !
! ! !! ! !!!!!! ! !
!!
! ! !!!!!!!!!!! ! !
>#!!!Z('(%;&.(!')(!0-<,'&-.427!H)(%(!!!!!!&2!.-'!8&//(%(.'&,?0(!,.8!H),'!/(,'+%(!;,I(2!&'!.-'!8&//(%(.'&,?0(#!
!!!!!!!!!!!!!!!!!!!!!! ! ! !!
! ! ! ! ! !!!!!!!!! ! ! ! ! ! ! !
!!!
!
!!!
5#!!!GI('<)!,!3%,1)!-/!/4@7!&/!! ! ! !
!
!!
!!!
!!!!,.8!! !! ! !!!
!!!
4. Draw a sketch of a graph that meets the following requirements. The limit does not exist as x
approaches -2. The functional value at x = 3 is equal to the limit of f as x approaches 3 but the function
is not differentiable at x = 3. The function increases without bound as x decreases without bound.!
:#!!!Z('(%;&.(!')(!=,0+(2!-/!R!,.8!S!'),'!;,I(!')(!/+.<'&-.!<-.'&.+-+2!,.8!8&//(%(.'&,?0(!,'!@!D!9>N!
! ! !
!! ! ! !! ! !! ! ! !!
!
!" ! !!
! ! !!
B#!!!Y('!!!! ! ! ! ! ! !#!!$)&<)!-/!')(!/-00-H&.3!2','(;(.'2!,%(!'%+(!,?-+'!"q!
K#!!!/!&2!<-.'&.+-+2!,'!! ! !!!
KK#!!!/!&2!8&//(%(.'&,?0(!,'!! ! !!!
KKK#!!!/!),2!,!<-%.(%!,'!! ! !!!
S"W!!!K!-.0A!
SOW!!!KK!-.0A!
S+W!!!KKK!-.0A!
S?W!!!K!,.8!KKK!-.0A!
SFW!!!K!,.8!KK!-.0A!!!!
M#!!!!$)&<)!-/!')(!/-00-H&.3!&2!'%+(!,?-+'!')(!3%,1)!-/!!!! ! ! ! !!! !!!,'!!!! ! !!q!
S"W!!!K'!),2!,!<-%.(%#!
SOW!!K'!),2!,!<+21#!
S+W!!!K'!),2!,!=(%'&<,0!',.3(.'#!
S?W!!!K'!),2!,!8&2<-.'&.+&'A#!!
SFW!!!! ! !8-(2!.-'!(@&2'!
O#!!!GI('<)!,!3%,1)!-/!,!<-.'&.+-+2!/+.<'&-.!!!!!!
!!!
! ! !!
&/!! ! ! !!!!!,.8N! ! ! ! ! !! !! ! ! ! !!
!!!
!!!
Q#!!!\)(!/+.<'&-.!)4@7!&2!3%,1)(8!?(0-H#!!
GI('<)!,!3%,1)!-/!')(!8(%&=,'&=(!-/!)4@7#!
y
4
y
3
6
2
5
1
4
3
–4
2
–5
–4
–3
–2
–1
–1
–2
–3
–4
–5
–6
–2
–1
1
–1
1
–6
–3
–2
1
2
3
4
5
6
x
–3
–4
2
3
4
x
+,-.%/0!P[!R!?=77%'%-0=(0=,-!N8#%1!
"#$%&!'!!
"#!![+0(2!/-%!*+&<I0A!8&//(%(.'&,'&.3r#!
,#!!!`%-8+<'!/+.<'&-.2!
?#!!!s+-'&(.'!/+.<'&-.2!
<#!!!\%&3-.-;('%&<!/+.<'&-.2!
8#!!!^@1-.(.'&,0!/+.<'&-.2!
(#!!Y-3,%&');&<!/+.<'&-.2!
/#!!!K.=(%2(!'%&3-.-;('%&<!/+.<'&-.2!
($)*+!'!+)),.//234)415#/4`LC>JGWaKEA!+)),.//234)415#/IMHOP@<_6[b!!
G%2!'!!
^=(%A!/+.<'&-.h2!8(%&=,'&=(!<,.!?(!/-+.8!+2&.3!')(!8(/&.&'&-.!-/!8(%&=,'&=(!!"#!!!
! !!! !! !
!!! !!
#!!]-H(=(%!')&2!&2!
'(8&-+2!,.8!%(1('&'&=(!'-!8-!-=(%!,.8!-=(%!,3,&.#!!S(0-H!&2!,!n1%--/o!-/!')(!8(%&=,'&=(!-/!')(!/+.<'&-.!! ! !
!"# !#!!!
Derivative proof of sin(x)
For this proof, we can use the limit definition of the derivative.
Limit Definition for sin:
Using angle sum identity, we get
Rearrange the limit so that the sin(x)'s are next to each other
Factor out a sin from the quantity on the right
Seperate the two quantities and put the functions with x in front of the limit (We are only concerned with the limit of h)
We can see that the first limit converges to 1 and the second limit converges to 0.
We can plug in 1 and 0 for the limits and get cos(x)
`]^$i!!X&=(.!')(!8&//&<+0'A!-/!'),'6!&'!?(<-;(2!1%+8(.'!'-!;(;-%&P(!')(!8(%&=,'&=(2!-/!<-;;-.!
/+.<'&-.2#!!`0(,2(!<-;;&'!'-!;(;-%A!')(!0&2'!?(0-H#!!C-%!2'+8A!)(01!1+%1-2(26!,!G\VZcSYV^!/0,2)!<,%8!
<-00(<'&-.!&2!,=,&0,?0(!,'!)''1Nkk2#'+8A#&'k'H2H,*:5!
d n
x = nx n!1
dx
d
c=0
dx
!!
d
[uv ] = u' v + uv'
dx
!
d ' u $ vu '!uv '
=
dx %& v "#
v2 !
d
sin x = cos x !
dx
d
tan x = sec 2 x !
dx
d
sec x = sec x tan x
dx
!
d
cos x = ! sin x
dx
!
d
cot x = ! csc 2 x
dx
!
d
csc x = ! csc x cot x !
dx
d x
e = ex
dx
!
d x
a = a x ln a !
dx
d
1
ln x = !
dx
x
!
!
!
d
1
!
log a x =
dx
x ln a
d
1
sin !1 x =
dx
1! x2 !
d
1
tan !1 x =
dx
1+ x2 !
1
d
sec !1 x =
dx
x x 2 !1
d
!1
cos !1 x =
dx
1! x2 !
d
!1
cot !1 x =
dx
1+ x2 !
!1
d
csc !1 x =
dx
x x 2 !1
Using the differentiation rules for all of the functions on the previous page, calculate 𝑓′(𝑥) for each of the following: 1. y = 7 − x
2. y = 6x2 – 3x + 2
3. y = x + 6
4. y =
5. y = 3 tan x
6. y = 4 x+3 7. 𝑦 = 12sec !! (𝑥)
8. y =
x
7
1
x 3 Derivatives involving the product and quotient rules
9. 𝑦 = csc 𝑥 − 𝑥 sin 𝑥
11. y =
6
tan x
10. y =
x2
cos x
12. y = x 5 ln x
13. u and v are differentiable functions such that 𝑢 5 = −3, 𝑣 5 = 1, 𝑢 ! 5 = 2, 𝑎𝑛𝑑 𝑣 ! 5 = −4. a. If 𝑎 𝑥 =
!(!)
!(!)
, find 𝑎 ! 5 . b. If 𝑏 𝑥 = 𝑢(𝑥) ∙ 𝑣(𝑥), find 𝑏 ! 5 . c. If 𝑐 𝑥 = 3𝑢 𝑥 + 2𝑣 𝑥 , find 𝑐 ! 5 Concept #6 – Position / Velocity / Acceleration Learn – 1. How position, velocity, and acceleration are related. 2. How to determine velocity and acceleration when given a position function. Watch – http://youtu.be/pFeuGMMiZWw Try – The relationships between position, velocity, and acceleration are the most common application of differential calculus (WELL HAI PHYSICS C!). A brief description of key terms is below: Position (𝑠(𝑡)) is the location of a particle at a given time. Velocity is the rate of change of position relative to change in time, so instantaneous velocity is the derivative of position (𝑣 𝑡 = 𝑠′(𝑡)). Acceleration is the rate of change of the velocity of an object, so instantaneous acceleration is the derivative of velocity, and thus the second derivative of position (𝑎 𝑡 = 𝑣 ! 𝑡 = 𝑠′′(𝑡)). Displacement is the net change in position, while distance is the total change in position. For example, if you want 10 feet, turn around and walk 4 feet back, your displacement is 6 feet while your distance travelled is 14 feet. Speed is the absolute value of velocity (𝑠𝑝𝑒𝑒𝑑 = |𝑣|. (Speed typically isn’t abbreviated with a single letter.) (The speedometer on your car doesn’t say “-­‐40 mph” when you drive back home.) 1. A particle moves along a line so that its position at any time 𝑡 ≥ 0 is given by the function 𝑠 𝑡 = 𝑡 ! − 9𝑡 ! + 15𝑡 + 3 where s is measured in meters and t is measured in seconds, for 𝑡 ≥ 0 . a. Find the particles position at time t = 3 seconds. b. Find the displacement during the first 3 seconds. c. Find the average velocity during the first 3 seconds. d. Find the instantaneous velocity when 𝑡 = 2. e. Find the time at which the acceleration is zero. f. At what time(s) does the particle stop? (This about what it means to be stopped in terms of p/v/a). 2. The figure to the right shows the velocity, 𝑣 𝑡 , in meters per second, of a body moving along a line. a. Sketch a graph of the acceleration, on the interval [ 0 , 8 ]. y
5
4
3
2
1
b. During what time intervals is the particle moving in the –1
1
2
3
4
5
6
7
–1
positive direction? –2
–3
–4
c. Deep Thought: Does the particle end up to the left or –5
right of where it starts? Use the graph of 𝑣 (𝑡) and give a concise but thorough explanation to support your answer. 8
9
x
Wrap-­‐up Problems: 1. Which of the following is (A) !
!!! !
!
!!!
!" !!!
? (B) 0 (C) −
! ! !!
!!
(D) 2𝑥 −
!
!!
− 1 (E) −
!
!!! !
!
2. Let 𝑓 𝑥 = 𝑥 − . Find 𝑓 ! (𝑥) . (A) 1 +
!
!!
!
(B) 1 −
!
!!
!
!
!
!!
(C) − +
(D) 𝑥′ −
!
!!
(E) 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 3. Find the velocity at the time when the acceleration is zero given the particle position function: 𝑠 𝑡 = 𝑡 ! − 6𝑡 ! + 8𝑡 + 1 (A) -­‐6 (B) -­‐5 (C) -­‐4 (D) -­‐3 (E) -­‐2 !
4. If 𝑠 𝑡 = sec(𝑡), find the acceleration of the function at the time when 𝑡 = . !
(A) 2 (B) 2 2 5. If ℎ(𝑥) = cos 𝑥, find ℎ
!"#
!!
!
(C) 2 + 1 . 6. Evaluate lim!→!
!"#
!
!!
!
!
!!"#
!
!
. (D) − 2 (E) 4 !
7. Find an equation of the tangent line to 𝑦 = sin 𝑥 + cos 𝑥 at 𝑥 = . !
!!!
8. If 𝑦 = csc 𝑥, find ! . !!
! !
9. Given 𝑟 𝑥 = 𝑓 𝑥 𝑔 𝑥 and 𝑝 𝑥 =
!
!
, evaluate each of the following using the graph below: a. 𝑟′(1) b. 𝑝′(8) c. 𝑟′(4) d. 𝑝′(4) 12. The coordinates s(t) of a moving body for various values of t are given. t (sec) s (ft) 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 -­‐19.5 -­‐6 4.5 12 16.5 18 16.5 12 4.5 a.) Assuming this smooth curve represents the motion of the body estimate the velocity at t = 1.0. b.) What must we know about s(t) in order to conclude that zeros must exist? Where would a zero have to exist if we knew that about s(t)?