Derivation of Some Differentiation Rules. f ( x + h ) f ( x ) lim. f ( x + h ) g( x + h ) f ( x ) g( x ) lim

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1 Derivation of Some Differentiation Rules Tese notes are intended to provide metods of deriving some of te formulas used in differentiation wic are different from tose described in te textbook We will be making use of te it definition ( -definition ) of te derivative of a function, f ' ( x ) f ( x + ) f ( x ) Since we can (and will often need to) construct new functions by combining simpler functions troug aritmetic operations or by composition of one function on anoter, we will need to know ow to differentiate tese newly-created functions Product Rule We can create a function F ( x ) f ( x ) g ( x ) troug multiplication of two simpler functions In calculating its derivative F ( x ), it will be convenient to define a symbol for te cange in a function by Δf f ( x + ) f ( x ), in order to save a bit of writing in places So we will ave f ( x + ) f ( x ) + Δf and we will need to apply binomial multiplication: F ' ( x ) [ f ( x ) g( x ) ]' f ( x + ) g( x + ) f ( x ) g( x ) [ f ( x ) + ] [ g( x ) + Δg ] f ( x ) g( x ) [ f ( x ) g( x ) + f ( x ) Δg + g( x ) + Δg ] f ( x ) g( x ) f ( x ) Δg + g( x ) + Δg At tis point, we can now express tis result as te sum of tree separate its and write out explicitly te canges in te functions f and g : [ f ( x ) g( x ) ]' f ( x ) Δg g( x ) + Δg + f ( x )[ g( x + ) g( x ) ] g( x )[ f ( x + ) f ( x ) ] + [ f ( x + ) f ( x ) ][ g( x + ) g( x ) ] +

2 g( x + ) g( x ) f ( x ) + g( x ) f ( x + ) f ( x ) + f ( x + ) f ( x ) [ g( x + ) g( x ) ], were we ave extracted te factor wic does not depend on in te first two of tese it terms, and ave simply separated one factor in te tird it By applying te it definition of a derivative function, we at last ave [ f ( x ) g( x ) ]' f ( x ) g ' ( x ) + g( x ) f ' ( x ) + f ( x + ) f ( x ) [ g( x + ) g( x ) ] te it of a product is te product of te its f ( x ) g ' ( x ) + g( x ) f ' ( x ) + f ' ( x ) [ g( x + 0) g( x ) ] * f ( x ) g ' ( x ) + g( x ) f ' ( x ) + f ' ( x ) 0 f ( x ) g ' ( x ) + g( x ) f ' ( x ) * We would obtain a similar result if we separated out te factor involving f instead More simply, te Product Rule is often expressed as ( f g ) f g + f g Notice tat tis Rule can be repeatedly applied to work out te derivative for a product of more tan two functions; for tree functions, for instance, ( f g )' ( f g )' + ( f g ) ' ( f ' g + f g ') + ( f g ) ' f ' g + f g ' + f g ' In oter words, te derivative of any product of functions can be expressed by a set of terms in wic eac function is differentiated in turn and multiplied by all te oter functions in te set Te Product Rule applies at tose values of x for wic every one of te functions in te product is continuous Quotient Rule We take a similar approac ere wit a new function defined by te ratio of two functions, G( x ) f ( x ) Naturally, we expect te algebra ere to be a little more g( x ) complicated G ' ( x ) f ( x ) ' g( x ) f ( x + ) g( x + ) f ( x ) g( x ) g( x ) f ( x + ) g( x + ) f ( x ) g( x + ) g( x ) subtracting fractions in te numerator g( x ) [ f ( x ) + ] f ( x ) [ g( x ) + Δg ] g( x + ) g( x )

3 g( x ) f ( x ) g( x ) + g( x ) f ( x ) g( x ) f ( x ) Δg g( x + ) using our expression for f ( x + ) and g ( x + ) and extracting a factor wic does not involve g( x ) g( x ) f ( x ) Δg g( x + ) We will now write out te canges in te functions f and g again, so tat we can apply te it definition of derivative: f ( x ) ' g( x ) g( x ) g( x + ) g( x ) [ f ( x + ) f ( x ) ] f ( x )[ g( x + ) g( x ) ] g( x ) g( x + ) g( x ) f ( x + ) f ( x ) f ( x ) g( x + ) g( x ) g( x ) g( x ) g( x + ) f ( x + ) f ( x ) extracting factors wic do not involve f ( x ) g( x + ) g( x ) g( x ) g( x ) f ' ( x ) f ( x ) g ' ( x ) g( x + 0) [ ] g( x ) f ' ( x ) f ( x ) g ' ( x ) [ g( x ) ] 2 Te Quotient Rule applies at tose values of x for wic bot f ( x ) and g ( x ) are continuous and were g ( x ) 0 (tat is, were f ( x )/g( x ) is defined and tus continuous) Cain Rule It is a bit more of a callenge to differentiate a composite function, wic is formed by taking te result of one function and subjecting it to te operation of a second function So we need to be somewat careful about wat te canges in te two functions mean Applying te it definition of derivative to te composite function H ( x ) f ( g ( x ) ), we ave H ' ( x ) [ f (g( x )) ] ' f ( g( x + )) f (g( x )) f ( g( x ) + Δg ) f (g( x )) We write te last expression in tis way as a reminder tat te cange in te composite function f ( g ( x ) ) is connected to te cange in te function g ( x ) Wen we ten use our way of sowing te sift in te value of te first term of te numerator to write

4 [ f (g( x )) ] ' [ f ( g( x )) + ] f (g( x )), it is ten peraps easier to keep in mind tat tis cange in te function f, Δf, is dependent upon te cange in te function g, Δg (wereas in our derivations of te Product and Quotient Rules above, tese canges were not connected) We can now say [ f (g( x )) ] ' f ( g( x )) + f (g( x )) Δg Δg Δg Δg Δg g( x + ) g( x ) Δg g ' ( x ) Wat remains to be understood is tis first it term Since it is certainly te case tat Δg approaces zero as approaces zero, we can tink of tis it as Δg Δg 0 Δg Δg 0 f ( g( x ) + Δg ) f (g( x )) Δg, reverting te numerator to a form it ad earlier But tis resembles te it definition for f ( x ), f ( x + ) f ( x ), wit Δg standing in for and g ( x ) in place of x Tis it in question ten gives te derivative function f ( u ) evaluated at te value u g ( x ) Tis permits us to write te Cain Rule for differentiation of a composite function, [ f (g( x )) ] ' Δg g' ( x ) f ' (u ) g' ( x ), u g(x) or, as it is often more simply written, [ f (g( x )) ] ' f ' (g( x )) g' ( x ) Te Cain Rule applies at tose values of x for wic bot g ( x ) and f ( g ( x ) ) are continuous

5 Derivatives of f ( x ) sin x and g ( x ) cos x Tese are te first of te elementary functions we encounter were someting more tan simple algebra is required in order to work out teir derivative functions We will need to construct a couple of new it laws for te purpose sin x Te first of tese is to find te value for One metod of x 0 x calculating tis is provided in te textbook (Stewart, 6 t ed, pp 90-9) A couple of oters are sown ere to offer alternative approaces For any of tese metods, we must consider a wedge of a circle of radius, wit center at point O and te angle AOB aving measure (size) Te area of tis wedge is A w ½ r 2 ½ 2 ½ We can extend a line downward from point A wic is perpendicular to te line OB and meets it at point C to form te rigt triangle ΔOCA From trigonometry, we know tat, since te ypotenuse OA is a radius of te circle and so as a lengt of, ten OC as lengt cos and AC as lengt sin Te segments OC and AC are te base and altitude of te rigt triangle ΔOCA, so its area is A OCA ½ cos sin We can ten also extend a line upward from point B wic is perpendicular to te line OB, and we will also extend te segment OA Tese lines meet at a point D, allowing us to make anoter rigt triangle ΔOBD Since OB is a radius of te circle, it as a lengt of Again, from trigonometry, te altitude of tis triangle BD as a eigt, tus / tan tan As te segments OB and BD are te base and altitude of tis rigt triangle, its area is A OBD ½ tan ½ tan Te wedge of te circle is enclosed between tese two rigt triangles, so we can write te inequality for te areas of tese geometrical figures as A OCA < A w < A OBD 2 cos sin < 2 < 2 tan If we now divide te inequality troug by ½ sin and take te it of te terms as te angle approaces zero, we ave

6 cos sin 2 2 sin < 2 2 sin < 2 tan 2 sin cos < sin < cos cos < sin < cos Upon evaluating te its at eac end of te inequality, we find < and terefore, by te Squeeze Teorem, laws, we can now write sin <, sin By anoter of te it sin sin, giving us our new trigonometric it law Anoter metod involves lengts of lines and arcs, rater tan te areas of wedges and triangles We start once again wit te wedge of te unit circle, OAB Since te angle AOB as measure, te lengt of te arc AB is s w r We again drop a perpendicular line from point A to te line OB to form te rigt triangle ΔOCA Tis time, we are interested in te lengt of tis line, wic is te altitude of te triangle we earlier found to be L AC sin We will now make a new circular wedge using te segment OC as te radius Te angle DOC must also ave measure We know tat OC as lengt cos, so te lengt of te arc CD is s CD r CD cos Te way in wic te altitude of te rigt triangle falls between te arcs of te two wedges gives us te inequality s CD < L AC < s w cos < sin <

7 We divide tis inequality troug by te angle and take te its of te terms as tis angle approaces zero: cos < sin < cos < sin < < sin <, wic gives us sin by te Squeeze Teorem We can proceed from tis result to te oter it law we will need We can make a product of certain its and ten use te already known it laws to write sin sin + cos sin sin + cos sin 2 ( + cos ) 0 ( cos 2 ) ( + cos ) 0 ( cos ) ( + cos ) 0 ( + cos ) applying te Pytagorean Identity factoring difference of two squares ( cos ) 0 safe to divide troug, since ( + cos ) 0 0 We now ave te trigonometric it laws we need to calculate te derivative functions for sin x and cos x Using te angle-addition formulas for sine and cosine (discussed in anoter Note), we ave te its [ sin x ]' sin ( x + ) sin x (sin x cos + cos x sin ) sin x (sin x cos sin x ) + (cos x sin ) sin x (cos ) + cos x sin sin x (cos ) + cos x sin (sin x 0) + (cos x ) cos x and

8 [ cos x ]' cos ( x + ) cos x (cos x cos sin x sin ) cos x (cos x cos cos x ) + ( sin x sin ) cos x (cos ) sin x sin cos x (cos ) sin x sin (cos x 0) (sin x ) sin x Derivative of te general exponential function Tis is anoter function tat requires some specific andling and also touces upon topics beyond te scope of Calculus I We can apply te it definition of derivative to te general exponential function f ( x ) a x, wit a > 0, to obtain [ a x ]' a x + a x (a x a ) a x a x applying properties of exponents a a x a extracting factor wic does not involve We are not in a position to evaluate tis last it (we will know ow to do tat in Calculus II), but we can recognize tat tis is te point derivative for our function, f ( 0 ), te slope of te tangent line to te exponential function y a x at x 0 (as discussed in Stewart, 6 t ed, pp 78-79) By experimenting wit different values of a > 0, we find tat tis it as a value wic depends upon te value of a Matematicians basically assign a name to te value at wic tis it is exactly ; tat number is called e (Tis is to say tat we don t prove tat e is te number for wic tis it is ; instead, we prove tat tere must be suc a number and te value at wic tis occurs is approximately , wic is designated as te constant e ) So we can say tat e and tus [ e x ]' e x e e x e x Te function e x is tus a function wic is its own derivative function; in fact, it is te only (non-constant) function for wic tat is te case Because it emerges directly from te structure of matematics, e x is called te natural exponential function

9 We can take tis a bit furter by looking at te function g ( x ) e kx, for wic te it definition of derivative yields [ e kx ]' e k ( x + ) e k x e k x e k, following te argument we used above for a x Now if k is a positive integer, we can write te numerator of te ratio in te it expression as ( e ) k, and apply te so-called geometric expression, x k ( x ) ( x k + x k 2 + K + x 2 + x + ), to re-write te derivative function as k terms [ e kx ]' e k x e k x (e ) k (e ) ([ e ] k + [ e ] k 2 + K + [ e ] 2 + e + ) e k x (e ) ([ e ] k + [ e ] k 2 + K + [ e ] 2 + e ) k terms e k x ([] k + [] k 2 + K + [] ) e k x k k terms Hence, we ave sown tat [ e kx ]' k e k x, at least wen k is a positive integer Tis is akin to te proof we ve given earlier in te course tat [ x n ]' n x n, were n is a positive integer (see, for example, Stewart, p 74) We can now sow immediately tat for a e k [ a x ]' k a x But from wat we ve learned prior to tis course,, wit k being a positive integer, tat a e k k ln a So we can argue plausibly tat [ a x ]' (ln a ) a x, even toug we ave really only so far sown it to be true wen ln a is a positive integer We will be able to demonstrate (elsewere) te derivative rule for a x more generally using te Cain Rule From te discussion earlier, we ave also sown tat te slope of a te tangent line to f ( x ) a x at x 0 is f ' (0) ln a -- G Ruffa May June 200