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NEWlogarithm

NEWlogarithm - The natural logarithm func1on ...

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Unformatted text preview: The natural logarithm func1on  Sec1on 7.2*  The natural logarithm func1on  •  •  •  •  •  •  Deﬁni1on  Graph  Proper1es  Deriva1ves that involve logarithm   Integrals that involve logarithm  Logarithm in diﬀerent bases  in base e=2.718281828…  The natural logarithm func1on  •  •  •  •  •  •  Deﬁni1on  Graph  Proper1es  Deriva1ves that involve logarithm   Integrals that involve logarithm  Logarithm in diﬀerent bases  Deﬁni1on of natural logarithm  For x>0, we deﬁne  Fix x>0. The func1on f(t)=1/t   is con1nuous over [1,x], so  this integral exists.  Natural logarithm in terms of area  f con1nuous over [a,b], and posi1ve  [a,b] is an interval  >a  b ∫ a f ( t ) dt = area beneath the graph of f , over [ a, b] € Natural logarithm in terms of area  con1nuous over [1,u], and posi1ve  for u>1, [1,u] is an interval     >1  Fix u>1.  1 A( u) = ∫ dt 1t u That’s what we deﬁned as ln(u)  Natural logarithm in terms of area  1 ln( x ) = ∫ dt = A( x ) for x ≥ 1 1t x € 1 1 ln( x ) = ∫ dt = − ∫ dt = − A( x ) for 0 < x < 1 1t xt x 1 € Sign of the Natural logarithm  ln( x ) = A( x ) for x ≥ 1   € ln(x)>0 for x>1, and  ln(x)=0 for x=1   ln( x ) = − A( x ) for 0 < x < 1   ln(x)<0 for 0<x<1  € 1 The natural logarithm ln( x ) = ∫ dt 1t in base e=2.718281828…  x € Ques1on  The shaded area is equal to:  A) ‐1/e2 +1  B) 1/e  C) 1  The natural logarithm func1on  •  •  •  •  •  •  Deﬁni1on  ✔  Graph  Proper1es  Deriva1ves that involve logarithm   Integrals that involve logarithm  Logarithm in diﬀerent bases  Graph of the func1on f(x)=ln(x)  • Domain: x>0  already  know:  0  1  x  • x‐intercept: x=1  • Posi1ve: x>1  • Nega1ve: 0<x<1  You need more informa2on:  •  behavior at end‐points of domain    •  deriva2ves (extremes etc..)   Graph of the func1on f(x)=ln(x)  Behavior at end‐point of domain:  x lim ln( x ) = lim ∫ x →∞ x →∞ 1 x 1 dt = +∞ t 1 lim ln( x ) = lim ∫ dt = −∞ x →0 x →0 1t 0  1  x  Graph of the func1on f(x)=ln(x)  DerivaAve:  d d ln( x ) = dx dx 1 1 ∫ t dt = x 1 x d ln( x ) > 0 for all x > 0 dx Fundamental theorem of calculus  € The func4on ln(x) is increasing for x>0.  It has no maxima and no minima.  Graph of the func1on f(x)=ln(x)  The natural logarithm func1on  •  •  •  •  •  •  Deﬁni1on  ✔  Graph  ✔  Proper1es  Deriva1ves that involve logarithm   Integrals that involve logarithm  Logarithm in diﬀerent bases  Proper1es of the logarithm     1)  ln(ab)=ln(a)+ln(b)  2)  ln(a/b)=ln(a)‐ln(b)  3)  ln(ar) = r ln(a), for all r ra1onal  Proper1es of the logarithm     1)  ln(ab)=ln(a)+ln(b)  By subs1tu1on: t=x/a  1/x      1/(at)  dx        a dt  a          1                ab        b   Proof:    ln(a)    Proper1es of the logarithm     2)  ln(a/b)=ln(a)‐ln(b)  Proof:  0 = ln(1) = ln(b/b) = ln(b(1/b)) = ln(b) + ln(1/b)   ln(1/b) = ‐ ln(b)    ln(a/b) = ln(a(1/b)) = ln(a) + ln(1/b) = ln(a) – ln(b)         Proper1es of the logarithm     3)  ln(ar) = r ln(a), for all r ra1onal  E.g.,  ln(a3)=ln(a3/2a3/2)=ln(a3/2)+ln(a3/2) = 2 ln(a3/2)      ln(a3/2)=(1/2) ln(a3)      ln(a3) = ln(aaa)= ln(a) + ln(a) + ln(a) = 3 ln(a)   ln(a3/2)=(1/2) ln(a3) = (1/2) 3 ln(a) = (1/2) 3 ln(a)        Proper1es of the logarithm     1)  ln(ab)=ln(a)+ln(b)  2)  ln(a/b)=ln(a)‐ln(b)  3)  ln(ar) = r ln(a), for all r ra1onal  Example:  ln ( x 2 + 4 ) 5 sin x x +1 2 3 = ln[( x 2 + 4 ) 5 sin x ] − ln [ x3 +1 = 1 = 5 ln( x + 4 ) + ln(sin x ) − ln( x 3 + 1) 2 Ques1on  The quan1ty                           is equal to  ln a(b 2 + c 2 ) ln( a) ln(b 2 + c 2 ) A) + €2 2 B) − 2 ln( a) − 2 ln(b + c ) ln( a) ln(2b + 2c ) + 2 2 2 2 C) The natural logarithm func1on  •  •  •  •  •  •  Deﬁni1on  ✔  Graph  ✔  Proper1es ✔  Deriva1ves that involve logarithm   Integrals that involve logarithm  Logarithm in diﬀerent bases  Deriva1ves of logarithm func1on  d 1 ln( f ( x )) = f '( x) dx f ( x) € ⏎  d d ln( x ) = dx dx 1 1 ∫ t dt = x 1 x Chain rule  Ques1on  The deriva1ve of the func1on f(x)=ln(tan(x)) is  A) 1 tan x B) 1 sec 2 x tan x 2 1 C ) sec x Ques1on  The deriva1ve of the func1on  f(x)=ln(x2sin2x)  is  A) 1 x 2 sin 2 x 1 1 +2 2 x sin x 2 2 cos x + x sin x B) C) Deriva1ves of logarithm func1on  d 1 ln( f ( x )) = f '( x) dx f ( x) € Hint: To simplify the calcula2ons, use the proper2es of the logarithm BEFORE diﬀeren2a2ng.  1)  ln(ab)=ln(a)+ln(b)  2)  ln(a/b)=ln(a)‐ln(b)  3)  ln(ar) = r ln(a), for all r ra1onal  •  Use the formula directly:   d 1 ln( f ( x )) = f '( x) dx f ( x) d ( x 2 + 4 ) 5 sin x 1 d ( x 2 + 4 ) 5 sin x ln =2 3 3 ( x + 4 ) 5 sin x dx dx x +1 x +1 € 3 x +1 f(x)  1/f(x)  f’(x)  RATHER  COMPLICATED!  •  Simplify ﬁrst:   ( x 2 + 4 ) 5 sin x d 1 2 3 ln = 5 ln( x + 4 ) + ln(sin x ) − 2 ln( x + 1) 3 x + 1 dx 1 1 11 MUCH EASIER!  =5 2 (2 x ) + (cos x ) − (3 x 2 ) 3 x +4 sin x 2 ( x + 1) d dx The natural logarithm func1on  •  •  •  •  •  •  Deﬁni1on  ✔  Graph  ✔  Proper1es ✔  Deriva1ves that involve logarithm ✔  Integrals that involve logarithm  Logarithms in diﬀerent bases  Logarithms and Integra1on  x if x > 0 | x |= − x if x > 0 ln( x ) if x > 0 ⇒ ln | x | = ln(− x ) if x > 0 d 1 if x > 0 dx ln( x ) if x > 0 x d ⇒ ln | x | = = = d 1 dx ln(− x ) if x > 0 (−1) if x > 0 dx −x 1 x ∫ 1 dx = ln | x | + C x   Logarithms and Integra1on  ∫ 1 dx = ln | x | + C x Then the subs1tu1on rule gives:  € 1 1 ∫ f ( x ) f ' ( x ) dx = ∫ u du = ln | u | + C = ln | f ( x ) | + C u=f(x)  Example:  ∫ tan x dx = ∫ 1 cos x dx = sin x ∫ 1 du = ln | u | + C = ln | sin x | + C u Ques1on  (ln x ) 2 The integral                      is solved with the subs1tu1on  ∫ x dx € A) u = 1 / x B) u = ln x C ) u = (ln x ) 2 The natural logarithm func1on  •  •  •  •  •  •  Deﬁni1on  ✔  Graph  ✔  Proper1es ✔  Deriva1ves that involve logarithm ✔  Integrals that involve logarithm ✔  Logarithms in diﬀerent bases  Change of base law  •  The natural logarithm ln(x) means log in base e  •  e is an irra1onal number ≈ 2.718281828  log e x ln x •  If a≠1 and a>0,  log a x = = log e a ln a Deriva1ve of y=logax  1/x  ...
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