Logarithmic, Eponential an Other Transcenental Fnctions 5: The Natral Logarithmic Fnction: Differentiation The Definition First, yo mst know the real efinition of the natral logarithm: ln= t (where > 0) t Reminers ( ) log a b = log a+ log b a log = log a log b log log n n n n n n n n b ( a ) = b logn( a) logb = for any vali base b log n b b The Nmber e Eler's Nmber; The Natral Nmber: e= lim + Also, it is the nmber so that e t= t The Derivative of the Natral Logarithmic Fnction By the Secon Fnamental Theorem of Calcls: ln( ) ln( f ( ) ) f = f ( ) ( ) = The chain rle version: Since the omain of the natral log is positive reals, we can sometimes rn into troble ths, there is another form: ln = This can help! Logarithmic Differentiation Sometimes, the se of logarithms can help yo to take erivatives that are otherwise mighty gly (by taking the log of both sies of the eqation) This is becase of the fantastic property of logarithms that allows yo to bring eponents own as coefficients Also, the proct/qotient properties come in hany The only catch is that in sing techniqe, yo'll technically nee to se implicit ifferentiation, which means that yo might have some y terms that nee to be epane with the original efinition HOLLOMAN S IB HL YEAR IB HL NOTES 05, PAGE OF 8
Eamples [] Fin g ( ) if g( ) log 0 ( ) ln( ) g( ) = log0( ) = g ( ) ln( 0) = = = ln 0 ln 0 ( ) ( ) [] Fin f ( ) if f ( ) = + y= ln( y) = ln = + + + ln( y) ln ( ) ln( ) = y + sbstitte: y = y = = + + + + + ln( y) ln( ) ln( ) = + Now take the erivative! y ( ) ( ) 5: The Natral Logarithmic Fnction: Integration The Integrals Solve an Jst work backwars! = ln + C Natrally, life is not always this simple; perhaps a version reay-mae for sbstittion is on orer: = ln + C There is not, however, an easy soltion to ln Perhaps we'll chat abot that later Integrals of Trigonometric Fnctions The integrals of sine an cosine are easy; what abot tangent? sin sin tan= = cos = cos, if yo se = cos Ths, tan= ln cos + C Yo can se a similar metho to fin cot= ln sin + C, sec= ln sec+ tan + C, an csc= ln csc+ cot + C Eamples []? = + HOLLOMAN S IB HL YEAR IB HL NOTES 05, PAGE OF 8
I notice that this is "one over something," an (natrally) think of logarithms If = +, then = Ths, = = + Aha! ln ln C C = + = + + [4] =? ln Sneaky at first glance, there are several things that I col se for however, leaving the log in the enominator isn't an option; that mst be part of my = ln = = So = = = = ln Finally, ln ln ln C C = + = + [5] sec =? Easy! sec = sec= ln sec+ tan + C = ln sec + tan + C csc [6] =? cot Another tricky one however, note that the erivative of the enominator is almost the same as csc the nmerator = cot = csc So = = ln + C = ln cot + C cot 54: Eponential Fnctions: Differentiation an Integration The Natral Eponential Fnction First, the real efinition: if f ( ) = ln, then e f ( ) = In other wors, y= e = lny The Natral Eponential fnction is efine as the inverse fnction of the natral logarithmic fnction Reminers m n m n e e = e + m e m n = e n e Don't forget that the omain of the natral log is the range of the natral eponential, etc Derivatives of Eponential Fnctions Since ln( e ) =, ln( e ) = [ ] This becomes e e e e = =! HOLLOMAN S IB HL YEAR IB HL NOTES 05, PAGE OF 8
If yo want to take the erivative of an eponential fnction other than the natral, yo'll have to se logarithmic ifferentiation more on that in the net section! Integrals of Eponential Fnctions Easy! e e = + C Eamples [7] Fin the graient of the line tangent to y e = at the point (,e ) First, the erivative: y = e At the inicate point, the graient is e e e [8] =? e + e Clearly, this reqires a sbstittion bt which piece? I sspect the enominator e e = e + e = ( e e ) So = = ln + C= lne + e + C e + e 55: Bases Other than e Definitions a e ( ) ln a = for any base (0,) (, ) log a a ln = = ln for any vali base a lna lna Differentiation an Integration a = [ log ] a ( ln a )( a ) = ( lna)( ) Eamples [9] Fin y if y =, an e = a = a + C lna First, take the log of both sies: ln( ) ln( ) y = Now pll that eponent own: ln( y) ln( ) = Remember that ln() is a constant! Take the erivative, remembering that yo are really oing implicit ifferentiation: y = ln( ) Solve for y : y = y ln( ) Finally, replace the y with its y original efinition in this case, y = : y = ln( ) [0] Fin f ( ) if f ( ) = log 0 Or, se the efinition above HOLLOMAN S IB HL YEAR IB HL NOTES 05, PAGE 4 OF 8
ln ln0 Change of base, anyone? f ( ) =, so f ( ) [] =? If =, then ( ln )( ) = So ln = + C= + C Or yo know ln ln = = Or, se the formla ln0 ln0 = Keep going! ln 57: Differential Eqations: Separation of Variables Definition A ifferential eqation is an eqation that involves erivatives Sons simple, eh? General an Particlar Soltions The soltion to a ifferential eqation is a fnction that, when it an its erivatives are plgge into the ifferential eqation, cases the eqation to be tre Sons complicate, eh? Working from a erivative to a fnction will (natrally) involve integration Since inefinite integration always creates a constant C, we may obtain general soltions (families of soltions that iffer only in the vale of the constant) As we have seen previosly, if some information (initial conition) is given, then we can fin a particlar soltion Differential eqations are classifie accoring to the highest erivative that appears in the eqation this is calle the orer of the ifferential eqation Separation of Variables If it is possible to write a ifferential eqation so that every term involving is on one sie of the eqation, an every term involving y is on the other, then the ifferential eqation is separable As a reslt, the eqation can be solving by separating the variables an integrating Homogenos Differential Eqations Some non-separable ifferential eqations can be mae separable throgh a change of variables however, this is more ifficlt, an not part of the IB syllabs (nor the AP, as best I can tell ), so we won't consier it Eamples [] Fin the general soltion to y 4 = Well the variables are alreay separate! Let's jst write it in ifferential form: y= ( 4) Now, integrate both sies: y= ( 4) y= 4 + C Note 4 that yo only nee one constant the two that yo wol normally get can be joine together into one HOLLOMAN S IB HL YEAR IB HL NOTES 05, PAGE 5 OF 8
[] Fin the general soltion to y = y First, let's change notation, an pt things in ifferential form: y= y Now, separate the variables: y= Integrate: ln y = ln + C Now, solve for y: y ln+ C C ln C e e e y = e = = Remember that absolte vale can be remove with a pls-mins symbol C y=± e [4] Fin the particlar soltion to y ln y = = if ( ) y ln ln ln Well, let's fin the general soltion first: = = = Now, ifferential form: ln y= Integrating the right sie reqires a sbstittion: = ln = Ths, the ifferential can be written y=, which integrates to y= + C Uno the sbstittion to get the general soltion: y= ( ln ) + C, works; ths, Now, for the particlars: y ( ) = is a way of saying that the point ( ) ln 0 = ln + = ( ) + C = ( ) + C = C So, the particlar soltion is y ( ) 58: Inverse Trigonometric Fnctions: Differentiation Definitions Recall that in general, none of the trigonometric fnctions have inverses, since none of them are one-to-one However, also recall that we (yo!) previosly restricte the omains of some of these fnctions in orer to create inverse fnctions Inverse Fnction Domain Range arcsine [-, ] π π, arccosine [-, ] [0, π] arctangent R π π, Or tet goes so far as to also create inverse fnction for the other three trigonometric fnctions! Of corse, there is rarely sch a nee especially since they are not covere in the IB crriclm HOLLOMAN S IB HL YEAR IB HL NOTES 05, PAGE 6 OF 8
Derivatives Yo can erive the erivatives of the inverse trigonometric fnctions throgh implicit ifferentiation [ arcsin] =, [ arccos] =, an [ arctan] = + Eamples [5] Fin arctan Basic rle an chain rle or jst follow the formla above arctan = = + + ( )( ) [6] Fin f ( ) if f ( ) arcsin Proct rle? ( ) = f = + arcsin 59: Inverse Trigonometric Fnctions: Integration Of corse, some fnctions, when integrate, give the inverse trigonometric fnctions! arcsin = + C, a a arccos C a = a +, an = arctan + C a + a a The ifference in the integrals of arcsine an arccosine are sch that arcsine is almost always se As was the case with natral logarithms, fining arcsin is mch harer, an will come later Eamples [7] Fin 4 This looks a lot like the pattern for arcsine bt with a sbstittion = = Ths, = = Contine: 4 arcsin( ) arcsin( ) = + C= + C HOLLOMAN S IB HL YEAR IB HL NOTES 05, PAGE 7 OF 8
cos [8] Fin + sin This looks like it might be an arctangent problem = sin = cos Ths, cos = + sin Aha! arctan( ) arctan( sin ) + = + C= + C + HOLLOMAN S IB HL YEAR IB HL NOTES 05, PAGE 8 OF 8