math hands
math hands Big & Famous Ideas Famous Limits Famous Derivatives sin(x) =1 x x lim (.998) = 0 x→∞ x 1 =e lim 1 + x→∞ x lim
x→0
1 − cos(x) =0 x x lim (1.003) = ∞ x→∞ x k lim 1 + = ek x→∞ x lim
[k]′ = 0 h i′ xk = kxk−1
x→0
[x]′ = 1 1 ∗ [ln(x)]′ = x h i′ [stuf f ]′ = eln(stuf f )
[ex ]′ = ex ′
[sin(x)] = cos(x)
[cos(x)] = − sin(x)
2
′
[tan(x)] = sec (x)
Famous Limit Ideas
[sec(x)]′ = sec(x) tan(x) ′
lim [k] = k
(Lk)
lim [x] = c
(Lx)
x→c x→c
lim [kf (x)] = k lim [f (x))
x→c
lim [f (x) + g(x)] = lim [f (x)) + lim (g(x)]
(LS=SL*)
lim [f (x) · g(x)] = lim [f (x)) · lim (g(x)] x→c x→c f (x) limx→c [f (x)) lim = x→c g(x) limx→c (g(x)] h i lim [f (g(x))] = f lim g(x)
(LP=PL*)
x→c
x→c
x→c
x→c
[cot(x)]′ = − csc2 (x)
[csc(x)]′ = − csc(x) cot(x)
[cosh(x)]′ = sinh(x) −1 [arccot(x)]′ = 2 x +1 −1 [arccos(x)]′ = √ 1 − x2
(Pk*)
x→c
x→c
[sinh(x)] = cosh(x) 1 [arctan(x)]′ = 2 x +1 1 [arcsin(x)]′ = √ 1 − x2
∗∗
′
Derivatives chain rule edition
(LQ=QL*) (limTcont**)
x→c
lim f (x)= L ⇐⇒
* need x > 0 ** stuf f can be any real function g(x) > 0
i d h d (stuf f )k = k(stuf f )k−1 · [stuf f ] dx dx 1 d d [ln(stuf f )] = · [stuf f ] dx stuf f dx h i d d estuf f = estuf f · [stuf f ] dx dx d d [sin(stuf f )] = cos(stuf f ) · [stuf f ] dx dx d d 1 [arctan(stuf f )] = · [stuf f ] dx (stuf f )2 + 1 dx d d [sec(stuf f )] = sec(stuf f ) tan(stuf f ) · [stuf f ] dx dx
(THE definition)
x→c
∀ǫ > 0∃δ > 0: 0 < |x − c| < δ =⇒ |f (x) − L| < ǫ *when each exists, is finite, and denominators not zero, k and c are real numbers **when limx→c g(x) = L each exists, is finite, and f (x) is continuous at x = L
Famous Indeterminate Forms 00
∞0
1∞ 0 0 ∞−∞
∞·0 ∞ ∞ sin(∞)
(*)
(*) (**) (*) (*) (*) (*)
* stuf f can be any real function g(x), when each derivative exists and is finite ** stuf f can be any real function g(x) > 0, when each derivative exists and is finite
*last one is not that famous
famous derivative ideas
Big Deal on Continuous (f + g)′ = f ′ + g ′
•
h
lim [f (g(x))] = f lim g(x)
x→c
x→c
i
(f · g)′ = f ′ · g + f · g ′
• if f and g are continuous at some point x = c then f + g,
f · g,
f ′ (c)= lim
x→c
f /g, kf + cg, are also continuous
• Intermediate Value Theorem
(kf )′ = k(f )′
f (x) − f (c) x−c
[f (g)]′ = f ′ (g)g ′ ′ f ′ · g − f · g′ f = g g2 f (c + h) − f (c) f ′ (c) = lim h→0 h
if each derivative exists and is finite blah blah..
• EVT =⇒ Rolles’s Thm =⇒ MVT =⇒ FTC
math hands
c
2007-2010 MathHands.com LLC
math hands Big & Famous Ideas Definition Identities Sum-Angle Identities opp hyp opp tan θ = adj sin θ tan θ = cos θ 1 csc θ = sin θ sin θ =
adj hyp adj cot θ = opp cos θ cot θ = sin θ 1 sec θ = cos θ
cos θ =
the M.O.T.A. 7−→
cos(a − b) = cos a cos b + sin a sin b
cos(a + b) = cos a cos b − sin a sin b sin(a + b) = sin a cos b + cos a sin b
sin(a − b) = sin a cos b − cos a sin b tan a + tan b tan(a + b) = 1 − tan a tan b tan a − tan b tan(a − b) = 1 + tan a tan b
Co-Function Identities Double-Angle Identities sin θ = cos (90◦ − θ)
tan θ = cot (90◦ − θ)
− cos θ = sin(θ − 90◦ )
cos θ = sin(90◦ − θ)
◦
sec θ = csc (90 − θ)
cos(2a) = cos2 a − sin2 a
◦
sin θ = cos (θ − 90 )
cos(2a) = 1 − 2 sin2 a 2 tan a tan(2a) = 1 − tan2 a 1 + cos 2θ cos2 θ = 2
Even & Odd Identities sin(−θ) = − sin θ
cos(−θ) = cos θ
tan(−θ) = − tan θ
cot(−θ) = − cot θ
csc(−θ) = − csc θ
1 [cos(a − b) − cos(a + b)] 2 1 cos a cos b = [cos(a − b) + cos(a + b)] 2 1 sin a cos b = [sin(a + b) + sin(a − b)] 2 sin a sin b =
Pythagoras Identities tan2 θ + 1 = sec2 θ
sin2 θ = 1 − cos2 θ
cos2 θ = 1 − sin2 θ
tan2 θ = sec2 θ − 1
cot2 θ = csc2 θ − 1 p tan θ = ± sec2 θ − 1
sin(2a) = 2 sin a cos a 1 − cos 2θ sin2 θ = 2
Product-to-Sums Identities
sec(−θ) = sec θ
sin2 θ + cos2 θ = 1
cos(2a) = 2 cos2 a − 1
Sums-to-Products Identities
cot2 θ + 1 = csc2 θ p sin θ = ± 1 − cos2 θ p sec θ = ± tan2 θ + 1
General Strategies to Prove Identities
sin a + sin b = 2 sin
a+b 2
cos
a−b 2
sin a − sin b = 2 sin
a−b 2
cos
a+b 2
a+b a−b cos 2 2 a−b a+b sin cos a − cos b = −2 sin 2 2
cos a + cos b = 2 cos
1. Work on sides independently. 2. Tweak a known Identity. 3. Look at Graphs. 4. Start with something amazingly
Euler’s Identity
creative and brilliant! eiθ = cos θ + i sin θ
math hands
c
2007-2010 MathHands.com LLC
math hands Derivatives Good for... Sir Newton Related Rates • [good for..] solving equations
• [when...] for given a differentiable function f (x), roughly speaking, when one can find a point, x1 near a root, c of f (x), then most of the time, repeated application of newton will take us closer to the solution, ”c” • [how...] x2 = x1 −
f (x1 ) f ′ (x1 )
L’Hopital • [good for..] L’Hopital is an excellent option to find limits.
• [good for..] describing rate of change of one quantity, A when the rate of change of another quantity B is known. The relationship between A and B must also be known. • [how...] example: suppose we know A2 + 5B = ln(B) then d simply take dt [] on each side, and plug in known quantities..and solve for wanted rate... A2 + 5B = ln(B) (given) d 2 d d A + 5B = (take dt []) [ln(B)] dt dt dA dB 1 dB 2A +5 = (now just plug in and solve) dt dt B dt
• [when...]
f (x) 0 = or x→c g(x) 0 f ′ (x) lim ′ exists x→c g (x) lim
Curve Sketching
±∞ ∞
(L’H form, and..) (then..L’H applies)
• [how..] lim
x→3
x2 − 9 2x = lim =6 x→3 1 x−3
(L’H...humm fancy huh..)
• [good for..] sketching curves
• [when...] for given function f (x), when one can study the sign of the first derivative of a function f ′ (x) one can tell where the function is increasing or decreasing since f ′ (c) > 0 implies f increasing near c and f ′ (c) < 0 implies f decreasing near c Key Idea intervals where derivative is positive and negative are separated by critical points, points where the derivative DNE or is zero. degree one sketch of f (x) 4 2
Linear Approximations −8 −6 −4 −2 −2 • [good for..] Approximating some values of some functions.
• [when...] When f (c) is difficult to evaluate, but the equation of the line tangent to f (x) near c is easy(-ier?). √ • [how..] example: approximate 3 132 √ √ First note 3 125 is easy to compute and near 3 132, so we √ 1 find the equation of the tangent line to f (x) = x 3 = 3 x: 1
f (x) = x 3
(given)
f (125) = 5 (we’ll need it..) −2 1 ′ (will need it.) f (x) = (x) 3 3 −2 1 1 f ′ (125) = (125) 3 = (will need it.) 3 75 ′ p(x) = f (125) + f (125)(x − 125) (standard eq of tan line) 1 p(x) = 5 + (x − 125) (plug in) 75 382 1 (132 − 125) = (plug in) p(132) = 5 + 75 75
thus
√ 3
132 ≈
f ′ (x)
2
4
6
−4
-b +++++++ - - ++++++ −6 −4 −2 0
2
Curve Sketching II
4
6
• [good for..] sketching curves
• [when...] for given function f (x), when one can study the sign of the first & second derivatives of a function one can tell concavity of f (x) since f ′′ (c) > 0 implies concave up near c and f ′′ (c) < 0 implies f concave down near c Key Idea intervals where second derivative is positive and negative are separated by critical points, points where the second derivative DNE or is zero.
382 75
math hands
c
2007-2010 MathHands.com LLC
Optimization
Mean Value Theorem
• [good for..] finding max or min value of a function
• [key idea...] according to Fermat Theorem if a differentiable function f (x) attains a local max/min at c then f ′ (c) = 0 thus to find local max/min we look to the places where f is not differentiable or has derivative zero.
• [says...] if f differentiable over [a, b] and a < b then.... there exists a point c ∈ (a, b) such that f (b) − f (a) b−a
f ′ (c) = • [key picture...]
• [general strategy to solve applied problems] 1. read: the problem until you can identify which function is to be optimized, call it f .
5
2. formula: determine a formula for f .
b
4
3. variables: if the found formula for f contains more than one variable, eliminate all but one.
3
4. seek critical pts: critical points (and endpoints) are candidates for local/absolute max/min
1
5. test each critical point can use second derivative, ex. f ′′ (c) > 0 means concave up, means local min.
f (x) b
2
−3
−2
−1 −1
b
1
2
3
4
−2
• [important...] by itself, like a very forgettable midwife, but it delivers BIGtime THE Fundamental Theorem of Calculus
Rolle’s Theorem • [says...] if f differentiable over [a, b] and a < b then.... IF f (a) = f (b) there exists a point c ∈ (a, b) such that
• [UBER-important edition] if F ′ (x) = f (x) then f (c)(b − a) = F (b) − F (a)
′
f (c) = 0 • [key picture...] b
5 4
f (x)
3 2 1 −3
−2
−1 −1
b
1
b
2
3
4
5
−2
• [importance...] it helps prove the Mean Value Theorem
math hands
c
2007-2010 MathHands.com LLC
x→0
1 − cos(x) =0 x x lim (1.003) = ∞ x→∞ x k lim 1 + = ek x→∞ x lim
[k]′ = 0 h i′ xk = kxk−1
x→0
[x]′ = 1 1 ∗ [ln(x)]′ = x h i′ [stuf f ]′ = eln(stuf f )
[ex ]′ = ex ′
[sin(x)] = cos(x)
[cos(x)] = − sin(x)
2
′
[tan(x)] = sec (x)
Famous Limit Ideas
[sec(x)]′ = sec(x) tan(x) ′
lim [k] = k
(Lk)
lim [x] = c
(Lx)
x→c x→c
lim [kf (x)] = k lim [f (x))
x→c
lim [f (x) + g(x)] = lim [f (x)) + lim (g(x)]
(LS=SL*)
lim [f (x) · g(x)] = lim [f (x)) · lim (g(x)] x→c x→c f (x) limx→c [f (x)) lim = x→c g(x) limx→c (g(x)] h i lim [f (g(x))] = f lim g(x)
(LP=PL*)
x→c
x→c
x→c
x→c
[cot(x)]′ = − csc2 (x)
[csc(x)]′ = − csc(x) cot(x)
[cosh(x)]′ = sinh(x) −1 [arccot(x)]′ = 2 x +1 −1 [arccos(x)]′ = √ 1 − x2
(Pk*)
x→c
x→c
[sinh(x)] = cosh(x) 1 [arctan(x)]′ = 2 x +1 1 [arcsin(x)]′ = √ 1 − x2
∗∗
′
Derivatives chain rule edition
(LQ=QL*) (limTcont**)
x→c
lim f (x)= L ⇐⇒
* need x > 0 ** stuf f can be any real function g(x) > 0
i d h d (stuf f )k = k(stuf f )k−1 · [stuf f ] dx dx 1 d d [ln(stuf f )] = · [stuf f ] dx stuf f dx h i d d estuf f = estuf f · [stuf f ] dx dx d d [sin(stuf f )] = cos(stuf f ) · [stuf f ] dx dx d d 1 [arctan(stuf f )] = · [stuf f ] dx (stuf f )2 + 1 dx d d [sec(stuf f )] = sec(stuf f ) tan(stuf f ) · [stuf f ] dx dx
(THE definition)
x→c
∀ǫ > 0∃δ > 0: 0 < |x − c| < δ =⇒ |f (x) − L| < ǫ *when each exists, is finite, and denominators not zero, k and c are real numbers **when limx→c g(x) = L each exists, is finite, and f (x) is continuous at x = L
Famous Indeterminate Forms 00
∞0
1∞ 0 0 ∞−∞
∞·0 ∞ ∞ sin(∞)
(*)
(*) (**) (*) (*) (*) (*)
* stuf f can be any real function g(x), when each derivative exists and is finite ** stuf f can be any real function g(x) > 0, when each derivative exists and is finite
*last one is not that famous
famous derivative ideas
Big Deal on Continuous (f + g)′ = f ′ + g ′
•
h
lim [f (g(x))] = f lim g(x)
x→c
x→c
i
(f · g)′ = f ′ · g + f · g ′
• if f and g are continuous at some point x = c then f + g,
f · g,
f ′ (c)= lim
x→c
f /g, kf + cg, are also continuous
• Intermediate Value Theorem
(kf )′ = k(f )′
f (x) − f (c) x−c
[f (g)]′ = f ′ (g)g ′ ′ f ′ · g − f · g′ f = g g2 f (c + h) − f (c) f ′ (c) = lim h→0 h
if each derivative exists and is finite blah blah..
• EVT =⇒ Rolles’s Thm =⇒ MVT =⇒ FTC
math hands
c
2007-2010 MathHands.com LLC
math hands Big & Famous Ideas Definition Identities Sum-Angle Identities opp hyp opp tan θ = adj sin θ tan θ = cos θ 1 csc θ = sin θ sin θ =
adj hyp adj cot θ = opp cos θ cot θ = sin θ 1 sec θ = cos θ
cos θ =
the M.O.T.A. 7−→
cos(a − b) = cos a cos b + sin a sin b
cos(a + b) = cos a cos b − sin a sin b sin(a + b) = sin a cos b + cos a sin b
sin(a − b) = sin a cos b − cos a sin b tan a + tan b tan(a + b) = 1 − tan a tan b tan a − tan b tan(a − b) = 1 + tan a tan b
Co-Function Identities Double-Angle Identities sin θ = cos (90◦ − θ)
tan θ = cot (90◦ − θ)
− cos θ = sin(θ − 90◦ )
cos θ = sin(90◦ − θ)
◦
sec θ = csc (90 − θ)
cos(2a) = cos2 a − sin2 a
◦
sin θ = cos (θ − 90 )
cos(2a) = 1 − 2 sin2 a 2 tan a tan(2a) = 1 − tan2 a 1 + cos 2θ cos2 θ = 2
Even & Odd Identities sin(−θ) = − sin θ
cos(−θ) = cos θ
tan(−θ) = − tan θ
cot(−θ) = − cot θ
csc(−θ) = − csc θ
1 [cos(a − b) − cos(a + b)] 2 1 cos a cos b = [cos(a − b) + cos(a + b)] 2 1 sin a cos b = [sin(a + b) + sin(a − b)] 2 sin a sin b =
Pythagoras Identities tan2 θ + 1 = sec2 θ
sin2 θ = 1 − cos2 θ
cos2 θ = 1 − sin2 θ
tan2 θ = sec2 θ − 1
cot2 θ = csc2 θ − 1 p tan θ = ± sec2 θ − 1
sin(2a) = 2 sin a cos a 1 − cos 2θ sin2 θ = 2
Product-to-Sums Identities
sec(−θ) = sec θ
sin2 θ + cos2 θ = 1
cos(2a) = 2 cos2 a − 1
Sums-to-Products Identities
cot2 θ + 1 = csc2 θ p sin θ = ± 1 − cos2 θ p sec θ = ± tan2 θ + 1
General Strategies to Prove Identities
sin a + sin b = 2 sin
a+b 2
cos
a−b 2
sin a − sin b = 2 sin
a−b 2
cos
a+b 2
a+b a−b cos 2 2 a−b a+b sin cos a − cos b = −2 sin 2 2
cos a + cos b = 2 cos
1. Work on sides independently. 2. Tweak a known Identity. 3. Look at Graphs. 4. Start with something amazingly
Euler’s Identity
creative and brilliant! eiθ = cos θ + i sin θ
math hands
c
2007-2010 MathHands.com LLC
math hands Derivatives Good for... Sir Newton Related Rates • [good for..] solving equations
• [when...] for given a differentiable function f (x), roughly speaking, when one can find a point, x1 near a root, c of f (x), then most of the time, repeated application of newton will take us closer to the solution, ”c” • [how...] x2 = x1 −
f (x1 ) f ′ (x1 )
L’Hopital • [good for..] L’Hopital is an excellent option to find limits.
• [good for..] describing rate of change of one quantity, A when the rate of change of another quantity B is known. The relationship between A and B must also be known. • [how...] example: suppose we know A2 + 5B = ln(B) then d simply take dt [] on each side, and plug in known quantities..and solve for wanted rate... A2 + 5B = ln(B) (given) d 2 d d A + 5B = (take dt []) [ln(B)] dt dt dA dB 1 dB 2A +5 = (now just plug in and solve) dt dt B dt
• [when...]
f (x) 0 = or x→c g(x) 0 f ′ (x) lim ′ exists x→c g (x) lim
Curve Sketching
±∞ ∞
(L’H form, and..) (then..L’H applies)
• [how..] lim
x→3
x2 − 9 2x = lim =6 x→3 1 x−3
(L’H...humm fancy huh..)
• [good for..] sketching curves
• [when...] for given function f (x), when one can study the sign of the first derivative of a function f ′ (x) one can tell where the function is increasing or decreasing since f ′ (c) > 0 implies f increasing near c and f ′ (c) < 0 implies f decreasing near c Key Idea intervals where derivative is positive and negative are separated by critical points, points where the derivative DNE or is zero. degree one sketch of f (x) 4 2
Linear Approximations −8 −6 −4 −2 −2 • [good for..] Approximating some values of some functions.
• [when...] When f (c) is difficult to evaluate, but the equation of the line tangent to f (x) near c is easy(-ier?). √ • [how..] example: approximate 3 132 √ √ First note 3 125 is easy to compute and near 3 132, so we √ 1 find the equation of the tangent line to f (x) = x 3 = 3 x: 1
f (x) = x 3
(given)
f (125) = 5 (we’ll need it..) −2 1 ′ (will need it.) f (x) = (x) 3 3 −2 1 1 f ′ (125) = (125) 3 = (will need it.) 3 75 ′ p(x) = f (125) + f (125)(x − 125) (standard eq of tan line) 1 p(x) = 5 + (x − 125) (plug in) 75 382 1 (132 − 125) = (plug in) p(132) = 5 + 75 75
thus
√ 3
132 ≈
f ′ (x)
2
4
6
−4
-b +++++++ - - ++++++ −6 −4 −2 0
2
Curve Sketching II
4
6
• [good for..] sketching curves
• [when...] for given function f (x), when one can study the sign of the first & second derivatives of a function one can tell concavity of f (x) since f ′′ (c) > 0 implies concave up near c and f ′′ (c) < 0 implies f concave down near c Key Idea intervals where second derivative is positive and negative are separated by critical points, points where the second derivative DNE or is zero.
382 75
math hands
c
2007-2010 MathHands.com LLC
Optimization
Mean Value Theorem
• [good for..] finding max or min value of a function
• [key idea...] according to Fermat Theorem if a differentiable function f (x) attains a local max/min at c then f ′ (c) = 0 thus to find local max/min we look to the places where f is not differentiable or has derivative zero.
• [says...] if f differentiable over [a, b] and a < b then.... there exists a point c ∈ (a, b) such that f (b) − f (a) b−a
f ′ (c) = • [key picture...]
• [general strategy to solve applied problems] 1. read: the problem until you can identify which function is to be optimized, call it f .
5
2. formula: determine a formula for f .
b
4
3. variables: if the found formula for f contains more than one variable, eliminate all but one.
3
4. seek critical pts: critical points (and endpoints) are candidates for local/absolute max/min
1
5. test each critical point can use second derivative, ex. f ′′ (c) > 0 means concave up, means local min.
f (x) b
2
−3
−2
−1 −1
b
1
2
3
4
−2
• [important...] by itself, like a very forgettable midwife, but it delivers BIGtime THE Fundamental Theorem of Calculus
Rolle’s Theorem • [says...] if f differentiable over [a, b] and a < b then.... IF f (a) = f (b) there exists a point c ∈ (a, b) such that
• [UBER-important edition] if F ′ (x) = f (x) then f (c)(b − a) = F (b) − F (a)
′
f (c) = 0 • [key picture...] b
5 4
f (x)
3 2 1 −3
−2
−1 −1
b
1
b
2
3
4
5
−2
• [importance...] it helps prove the Mean Value Theorem
math hands
c
2007-2010 MathHands.com LLC
