18.01 Single Variable Calculus Fall 2006

Transcription

1 MIT OpenCourseWare Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit:

2 Lecture 8.0 Fall 2006 Unit : Derivatives A. What is a erivative? Geometric interpretation Physical interpretation Important for any measurement (economics, political science, finance, physics, etc.) B. How to ifferentiate any function you know. ( ) For eample: e arctan. We will iscuss what a erivative is toay. Figuring out how to ifferentiate any function is the subject of the first two weeks of this course. Lecture : Derivatives, Slope, Velocity, an Rate of Change Geometric Viewpoint on Derivatives y Q Secant line P Tangent line f() Figure : A function with secant an tangent lines The erivative is the slope of the line tangent to the graph of f(). But what is a tangent line, eactly?

3 Lecture 8.0 Fall 2006 It is NOT just a line that meets the graph at one point. It is the limit of the secant line (a line rawn between two points on the graph) as the istance between the two points goes to zero. Geometric efinition of the erivative: Limit of slopes of secant lines P Q as Q P (P fie). The slope of P Q: Q ( 0 +, f( 0 + )) Secant Line f ( 0, f( 0 )) P Figure 2: Geometric efinition of the erivative lim Δf f( 0 + Δ) f( 0 ) = lim = f ( 0 ) Δ 0 Δ Δ 0 } Δ {{}}{{} ifference quotient erivative of f at 0 Eample. f() = One thing to keep in min when working with erivatives: it may be tempting to plug in Δ = 0 Δf 0 right away. If you o this, however, you will always en up with =. You will always nee to Δ 0 o some cancellation to get at the answer. Δf [ 0 ( 0 + Δ) ] [ ] = 0 +Δ 0 Δ = = = Δ Δ Δ ( 0 + Δ) 0 Δ ( 0 + Δ) 0 ( 0 + Δ) 0 Taking the limit as Δ 0, lim = Δ 0 ( 0 + Δ)

4 Lecture 8.0 Fall 2006 y 0 Figure 3: Graph of Hence, f ( 0 ) = 2 0 Notice that f ( 0 ) is negative as is the slope of the tangent line on the graph above. Fining the tangent line. Write the equation for the tangent line at the point ( 0, y 0 ) using the equation for a line, which you all learne in high school algebra: y y 0 = f ( 0 )( 0 ) Plug in y 0 = f( 0 ) = an f ( 0 ) = 2 to get: 0 0 y 0 = 2 0 ( 0 ) 3

5 Lecture 8.0 Fall 2006 y 0 Figure 4: Graph of Just for fun, let s compute the area of the triangle that the tangent line forms with the - an y-aes (see the shae region in Fig. 4). First calculate the -intercept of this tangent line. The -intercept is where y = 0. Plug y = 0 into the equation for this tangent line to get: 0 = 2 ( 0 ) 0 0 = = = 2 0( ) = So, the -intercept of this tangent line is at = 2 0. Net we claim that the y-intercept is at y = 2y 0. Since y = an = are ientical equations, y the graph is symmetric when an y are echange. By symmetry, then, the y-intercept is at y = 2y 0. If you on t trust reasoning with symmetry, you may follow the same chain of algebraic reasoning that we use in fining the -intercept. (Remember, the y-intercept is where = 0.) Finally, Area = (2y 0 )(2 0 ) = 2 0 y 0 = 2 0 ( ) = 2 (see Fig. 5) 2 0 Curiously, the area of the triangle is always 2, no matter where on the graph we raw the tangent line. 4

6 Lecture 8.0 Fall 2006 y 2y 0 - y Figure 5: Graph of Notations Calculus, rather like English or any other language, was evelope by several people. As a result, just as there are many ways to epress the same thing, there are many notations for the erivative. Since y = f(), it s natural to write Δy = Δf = f() f( 0 ) = f( 0 + Δ) f( 0 ) We say Delta y or Delta f or the change in y. If we ivie both sies by Δ = 0, we get two epressions for the ifference quotient: Taking the limit as Δ 0, we get Δy Δf = Δ Δ Δy Δ Δf Δ y (Leibniz notation) f ( 0 ) (Newton s notation) When you use Leibniz notation, you have to remember where you re evaluating the erivative in the eample above, at = 0. Other, equally vali notations for the erivative of a function f inclue f, f, an Df 5

7 Lecture 8.0 Fall 2006 Eample 2. f() = n where n =, 2, 3... What is n? To fin it, plug y = f() into the efinition of the ifference quotient. Δy = ( 0 + Δ) n 0 = ( + Δ)n Δ Δ Δ (From here on, we replace 0 with, so as to have less writing to o.) Since ( + Δ) n = ( + Δ)( + Δ)...( + Δ) n times We can rewrite this as ( ) n + n(δ) n + O (Δ) 2 O(Δ) 2 is shorthan for all of the terms with (Δ) 2, (Δ) 3, an so on up to (Δ) n. (This is part of what is known as the binomial theorem; see your tetbook for etails.) Take the limit: Therefore, n Δy = ( + Δ) n = n + n(δ)( n ) + O(Δ) 2 = n n + O(Δ) Δ Δ Δ n Δy lim = n n 0 Δ Δ n n n = n n This result etens to polynomials. For eample, 9 ( ) = Physical Interpretation of Derivatives You can think of the erivative as representing a rate of change (spee is one eample of this). On Halloween, MIT stuents have a traition of ropping pumpkins from the roof of this builing, which is about 400 feet high. The equation of motion for objects near the earth s surface (which we will just accept for now) implies that the height above the groun y of the pumpkin is: y = 400 6t 2 Δy istance travelle The average spee of the pumpkin (ifference quotient) = = Δt When the pumpkin hits the groun, y = 0, time elapse 400 6t 2 = 0 6

8 Lecture 8.0 Fall 2006 Solve to fin t = 5. Thus it takes 5 secons for the pumpkin to reach the groun. 400 ft Average spee = = 80 ft/s 5 sec A spectator is probably more intereste in how fast the pumpkin is going when it slams into the groun. To fin the instantaneous velocity at t = 5, let s evaluate y : y = 32t = ( 32)(5) = 60 ft/s (about 0 mph) y is negative because the pumpkin s y-coorinate is ecreasing: it is moving ownwar. 7

9 Lecture Fall 2006 Lecture 2: Limits, Continuity, an Trigonometric Limits More about the rate of change interpretation of the erivative y y = f() y Figure : Graph of a generic function, with Δ an Δy marke on the graph Δy Δ y as Δ 0 Average rate of change Instantaneous rate of change Eamples. q = charge 2. s = istance q t = electrical current s t = spee T 3. T = temperature = temperature graient

10 Lecture Fall Sensitivity of measurements: An eample is carrie out on Problem Set. In GPS, raio signals give us h up to a certain measurement error (See Fig. 2 an Fig. 3). The question is ΔL how accurately can we measure L. To ecie, we fin. In other wors, these variables are Δh relate to each other. We want to fin how a change in one variable affects the other variable. satellite s h L you Figure 2: The Global Positioning System Problem (GPS) s h L Figure 3: On problem set, you will look at this simplifie flat earth moel 2

11 Lecture Fall 2006 Easy Limits Limits an Continuity lim = = = With an easy limit, you can get a meaningful answer just by plugging in the limiting value. Remember, lim Δf = lim f( 0 + Δ) f( 0 ) 0 Δ 0 Δ is never an easy limit, because the enominator Δ = 0 is not allowe. (The limit compute uner the implicit assumption that 0.) 0 is Continuity We say f() is continuous at 0 when lim f() = f( 0 ) 0 Pictures y Figure 4: Graph of the iscontinuous function liste below { + > 0 f() = 0 3

12 Lecture Fall 2006 This iscontinuous function is seen in Fig. 4. For > 0, lim f() = 0 but f(0) = 0. (One can also say, f is continuous from the left at 0, not the right.). Removable Discontinuity Figure 5: A removable iscontinuity: function is continuous everywhere, ecept for one point Definition of removable iscontinuity Right-han limit: lim f() means lim f() for > Left-han limit: lim f() means lim f() for < 0. 0 If lim f() = lim f() but this is not f( 0 ), or if f( 0 + ) is unefine, we say the isconti nuity is removable. For eample, sin( is efine for = 0. We will see later how to evaluate the limit as 0. ) 4

13 Lecture Fall Jump Discontinuity 0 Figure 6: An eample of a jump iscontinuity lim for ( < 0 ) eists, an lim for ( > 0 ) also eists, but they are NOT equal Infinite Discontinuity y Figure 7: An eample of an infinite iscontinuity: Right-han limit: lim = ; Left-han limit: lim 0 = 0 + 5

14 Lecture Fall Other (ugly) iscontinuities Figure 8: An eample of an ugly iscontinuity: a function that oscillates a lot as it approaches the origin This function oesn t even go to ± it oesn t make sense to say it goes to anything. For something like this, we say the limit oes not eist. 6

15 Lecture Fall 2006 Picturing the erivative y y Figure 9: Top: graph of f () = an Bottom: graph of f () = 2 Notice that the graph of f() oes NOT look like the graph of f ()! (You might also notice that f() is an o function, while f () is an even function. The erivative of an o function is always even, an vice versa.) 7

16 Lecture Fall 2006 Pumpkin Drop, Part II This time, someone throws a pumpkin over the tallest builing on campus. Figure 0: y = 400 6t 2, 5 t 5 Figure : Top: graph of y(t) = 400 6t 2. Bottom: the erivative, y (t) 8

17 Lecture Fall 2006 Two Trig Limits Note: In the epressions below, θ is in raians NOT egrees! lim sin θ = ; lim cos θ = 0 θ 0 θ θ 0 θ Here is a geometric proof for the first limit: θ sinθ arc length = θ Figure 2: A circle of raius with an arc of angle θ θ sin θ arc length = θ Figure 3: The sector in Fig. 2 as θ becomes very small Imagine what happens to the picture as θ gets very small (see Fig. 3). As θ 0, we see that sin θ. θ 9

18 Lecture Fall 2006 What about the secon limit involving cosine? arc length = θ θ - cos θ cos θ Figure 4: Same picture as Fig. 2 ecept that the horizontal istance between the ege of the triangle an the perimeter of the circle is marke From Fig. 5 we can see that as θ 0, the length cos θ of the short segment gets much smaller than the vertical istance θ along the arc. Hence, cos θ 0. θ θ cos θ - cos θ arc length = θ Figure 5: The sector in Fig. 4 as θ becomes very small 0

19 Lecture Fall 2006 We en this lecture with a theorem that will help us to compute more erivatives net time. Theorem: Differentiable Implies Continuous. If f is ifferentiable at 0, then f is continuous at 0. [ ] f() f( 0 ) Proof: lim (f() f( 0 )) = lim ( 0 ) = f ( 0 ) 0 = Remember: you can never ivie by zero! The first step was to multiply by 0. It looks as 0 0 if this is illegal because when = 0, we are multiplying by. But when computing the limit as 0 0 we always assume 0. In other wors 0 0. So the proof is vali.

20 Lecture Fall 2006 Lecture 3 Derivatives of Proucts, Quotients, Sine, an Cosine There are two kins of erivative formulas: ( ). Specific Eamples: n or Derivative Formulas 2. General Eamples: (u + v) = u + v an (cu) = cu (where c is a constant) A notational convention we will use toay is: (u + v)() = u() + v(); uv() = u()v() Proof of (u + v) = u + v. (General) Start by using the efinition of the erivative. (u + v) () = lim (u + v)( + Δ) (u + v)() Δ 0 Δ (u + v) () = u () + v () Follow the same proceure to prove that (cu) = cu. u( + Δ) + v( + Δ) u() v() = lim Δ 0 { Δ } u( + Δ) u() v( + Δ) v() = lim + Δ 0 Δ Δ Derivatives of sin an cos. (Specific) Last time, we compute sin lim = 0 (sin ) =0 = lim sin(0 + Δ) sin(0) = lim sin(δ) = Δ 0 Δ Δ 0 Δ (cos ) =0 = lim cos(0 + Δ) cos(0) = lim cos(δ) = 0 Δ 0 Δ Δ 0 Δ So, we know the value of sin an of cos at = 0. Let us fin these for arbitrary. sin( + Δ) sin() sin = lim Δ 0 Δ

21 Lecture Fall 2006 Recall: sin(a + b) = sin(a) cos(b) + sin(b) cos(a) So, sin cos Δ + cos sin Δ sin() sin = lim Δ 0 [ Δ ] sin (cos Δ ) cos sin Δ = lim + Δ 0 Δ Δ ( ) ( ) cos Δ sin Δ = lim sin + lim cos Δ 0 Δ Δ 0 Δ Since cos Δ 0 an that sin Δ, the equation above simplifies to Δ Δ A similar calculation gives sin = cos cos = sin Prouct formula (General) (uv) = u v + uv Proof: (uv) = lim (uv)( + Δ) (uv)() u( + Δ)v( + Δ) u()v() = lim Δ 0 Δ Δ 0 Δ Now obviously, u( + Δ)v() u( + Δ)v() = 0 so aing that to the numerator won t change anything. (uv) = lim u( + Δ)v() u()v() + u( + Δ)v( + Δ) u( + Δ)v() Δ 0 Δ We can re-arrange that epression to get ( ) ( ) (uv) u( + Δ) u() v( + Δ) v() = lim v() + u( + Δ) Δ 0 Δ Δ Remember, the limit of a sum is the sum of the limits. [ ] ( [ ]) lim u( + Δ) u() v( + Δ) v() v() + lim u( + Δ) Δ 0 Δ Δ 0 Δ Note: we also use the fact that (uv) = u ()v() + u()v () lim u( + Δ) = u() Δ 0 (true because u is continuous) This proof of the prouct rule assumes that u an v have erivatives, which implies both functions are continuous. 2

22 Lecture Fall 2006 v v u u Figure : A graphical proof of the prouct rule An intuitive justification: We want to fin the ifference in area between the large rectangle an the smaller, inner rectangle. The inner (orange) rectangle has area uv. Define Δu, the change in u, by Δu = u( + Δ) u() We also abbreviate u = u(), so that u( + Δ) = u + Δu, an, similarly, v( + Δ) = v + Δv. Therefore the area of the largest rectangle is (u + Δu)(v + Δv). If you let v increase an keep u constant, you a the area shae in re. If you let u increase an keep v constant, you a the area shae in yellow. The sum of areas of the re an yellow rectangles is: [u(v + Δv) uv] + [v(u + Δu) uv] = uδv + vδu If Δu an Δv are small, then (Δu)(Δv) 0, that is, the area of the white rectangle is very small. Therefore the ifference in area between the largest rectangle an the orange rectangle is approimately the same as the sum of areas of the re an yellow rectangles. Thus we have: [(u + Δu)(v + Δv) uv] uδv + vδu (Divie by Δ an let Δ 0 to finish the argument.) 3

23 Lecture Fall 2006 Quotient formula (General) To calculate the erivative of u/v, we use the notations Δu an Δv above. Thus, u( + Δ) u() v( + Δ) v() = u + Δu u v + Δv v = (u + Δu)v u(v + Δv) (common enominator) (v + Δv)v (Δu)v u(δv) = (cancel uv uv) (v + Δv)v Hence, ( u + Δu Δ v + Δv u ) v = ( Δu Δv )v u( Δ Δ ) (v + Δv)v v( u v ) u( ) v 2 as Δ 0 Therefore,. ( u v ) = u v uv v 2 4

24 Lecture 4 Sept. 4, Fall 2006 Lecture 4 Chain Rule, an Higher Derivatives Chain Rule We ve got general proceures for ifferentiating epressions with aition, subtraction, an multi plication. What about composition? Eample. y = f() = sin, = g(t) = t 2. So, y = f(g(t)) = sin(t 2 ). To fin y, write t t 0 = t 0 0 = g(t 0 ) y 0 = f( 0 ) t = t 0 + Δt = 0 + Δ y = y 0 + Δy Δy Δy Δ = Δt Δ Δt As Δt 0, Δ 0 too, because of continuity. So we get: y t y = The Chain Rule! t In the eample, t y = 2t an = cos. So, t ( sin(t 2 ) ) = ( y )( t ) = = (cos )(2t) (2t) ( cos(t 2 ) ) Another notation for the chain rule ( ) t f(g(t)) = f (g(t))g (t) or f(g()) = f (g())g () Eample. (continue) Composition of functions f() = sin an g() = 2 (f g)() = f(g()) = sin( 2 ) (g f)() = g(f()) = sin 2 () Note: f g g f. Not Commutative!

25 Lecture 4 Sept. 4, Fall 2006 g g() f f(g()) ( ) Eample 2. cos =? Let u = Figure : Composition of functions: f g() = f(g()) y y u = u y u = sin(u); u = 2 ( ) ( ) sin y sin(u) = = ( sin u) = ( ) Eample 3. n =? ( ) n There are two ways to procee. n =, or n = n. 2. ( n ) ( ) n ( ) n ( ) = = n = n (n ) 2 = n n 2 ( ) ( ) ( n ) = = n n = n n (Think of n as u) n 2n 2

26 Lecture 4 Sept. 4, Fall 2006 Higher Derivatives Higher erivatives are erivatives of erivatives. For instance, if g = f, then h = g is the secon erivative of f. We write h = (f ) = f. Notations f () f () f () f (n) () Df D 2 f D 3 f D n f f 2 f 2 3 f 3 n f n Higher erivatives are pretty straightforwar - just keep taking the erivative! n Eample. D n =? Start small an look for a pattern. D = D 2 2 = D(2) = 2 ( = 2) D 3 3 = D 2 (3 2 ) = D(6) = 6 ( = 2 3) D 4 4 = D 3 (4 3 ) = D 2 (2 2 ) = D(24) = 24 ( = 2 3 4) D n n = n! we guess, base on the pattern we re seeing here. The notation n! is calle n factorial an efine by n! = n(n ) 2 Proof by Inuction: We ve alreay checke the base case (n = ). n Inuction step: Suppose we know D n = n! (n th case). Show it hols for the (n + ) st case. ( ) D n+ n+ = D n D n+ = D n ((n + ) n ) = (n + )D n n = (n + )(n!) D n+ n+ = (n + )! Prove! 3

27 Lecture Fall 2006 Lecture 5 Implicit Differentiation an Inverses Implicit Differentiation Eample. ( a ) = a a. We prove this by an eplicit computation for a = 0,, 2,... From this, we also got the formula for a =, 2,... Let us try to eten this formula to cover rational numbers, as well: m m a = ; y = n where m an n are integers. n We want to compute y. We can say y n = m so ny n y = m m. Solve for y : y m m = n y n We know that y = ( m n ) is a function of. ( ) y m m = n y n ( ) m m = n ( m/n ) n m = n m(n )/n m = (m ) m(n ) n n m n(m ) m(n ) = n n m nm n nm+m = n n m m m = n n n y m m So, = n n This is the same answer as we were hoping to get! Eample 2. Equation of a circle with a raius of : 2 +y 2 = which we can write as y 2 = 2. So y = ± 2. Let us look at the positive case: n y = + ( ) 2 = ( 2 ) 2 y = ( 2 ) 2 ( 2) = 2 2 = y

28 Lecture Fall 2006 Now, let s o the same thing, using implicit ifferentiation. Applying chain rule in the secon term, Eample y 2 = ( 2 ) 2 + y = () = 0 ( 2 ) + (y 2 ) = y y = 0 2y y = 2 y = y Same answer! y 3 + y 2 + = 0. In this case, it s not easy to solve for y as a function of. Instea, we use implicit ifferentiation to fin y. We can now solve for y in terms of y an. Inverse Functions 3y 2 y + y 2 + 2y y = 0 y (3y 2 + 2y) = y 2 y = y 2 3y 2 + 2y If y = f() an g(y) =, we call g the inverse function of f, f : = g(y) = f (y) Now, let us use implicit ifferentiation to fin the erivative of the inverse function. y = f() f (y) = (f (y)) = () = By the chain rule: y (f (y)) y = an (f (y)) = y y 2

29 Lecture Fall 2006 So, implicit ifferentiation makes it possible to fin the erivative of the inverse function. Eample. y = arctan() tan y = [tan(y)] = = y [tan(y)] y = ( ) y cos 2 = (y) y = cos 2 (y) = cos 2 (arctan()) This form is messy. Let us use some geometry to simplify it. y (+ 2 ) /2 Figure : Triangle with angles an lengths corresponing to those in the eample illustrating ifferentiation using the inverse function arctan In this triangle, tan(y) = so arctan() = y The Pythagorian theorem tells us the length of the hypotenuse: h = + 2 From this, we can fin From this, we get cos(y) = + 2 ( ) 2 cos 2 (y) = =

30 Lecture Fall 2006 So, In other wors, y = + 2 arctan() = + 2 Graphing an Inverse Function. Suppose y = f() an g(y) = f (y) =. To graph g an f together we nee to write g as a function of the variable. If g() = y, then = f(y), an what we have one is to trae the variables an y. This is illustrate in Fig. 2 f (f()) = f f() = f(f ()) = f f () = y f() y= g() b=f(a) a=f - (b) Figure 2: You can think about f as the graph of f reflecte about the line y = 4

31 Lecture Fall 2006 Lecture 6: Eponential an Log, Logarithmic Differentiation, Hyperbolic Functions Taking the erivatives of eponentials an logarithms Backgroun We always assume the base, a, is greater than. a 0 = ; a = a; a 2 = a a;... a +2 = a a 2 (a ) 2 2 = a p q a q = a p (where p an q are integers) r To efine a for real numbers r, fill in by continuity. Toay s main task: fin We can write a a +Δ a a = lim Δ 0 Δ We can factor out the a : Let s call +Δ Δ Δ lim a a = lim a a = a lim a Δ 0 Δ Δ 0 Δ Δ 0 Δ M(a) lim aδ Δ 0 Δ We on t yet know what M(a) is, but we can say Here are two ways to escribe M(a): a = M(a)a. Analytically M(a) = a at = 0. Inee, M(a) = lim a0+δ a 0 = a Δ 0 Δ =0

32 Lecture Fall 2006 a M(a) (slope of a at =0) Figure : Geometric efinition of M(a) 2. Geometrically, M(a) is the slope of the graph y = a at = 0. The trick to figuring out what M(a) is is to beg the question an efine e as the number such that M(e) =. Now can we be sure there is such a number e? First notice that as the base a increases, the graph a gets steeper. Net, we will estimate the slope M(a) for a = 2 an a = 4 geometrically. Look at the graph of 2 in Fig. 2. The secant line from (0, ) to (, 2) of the graph y = 2 has slope. Therefore, the slope of y = 2 at = 0 is less: M(2) < (see Fig. 2). Net, look at the graph of 4 in Fig. 3. The secant line from ( 2, 2 ) to (, 0) on the graph of y = 4 has slope. Therefore, the slope of y = 4 at = 0 is greater than M(4) > (see Fig. 3). Somewhere in between 2 an 4 there is a base whose slope at = 0 is. 2

33 Lecture Fall 2006 y=2 slope = (,2) secant line slope M(2) Figure 2: Slope M(2) < y=4 (-/2, /2) (,0) slope M(4) secant line Figure 3: Slope M(4) > 3

34 Lecture Fall 2006 Thus we can efine e to be the unique number such that or, to put it another way, or, to put it still another way, M(e) = lim eh = h 0 h (e ) = at = 0 What is (e )? We just efine M(e) =, an (e ) = M(e)e. So (e ) = e Natural log (inverse function of e ) To unerstan M(a) better, we stuy the natural log function ln(). This function is efine as follows: If y = e, then ln(y) = (or) If w = ln(), then e = w Note that e is always positive, even if is negative. Recall that ln() = 0; ln() < 0 for 0 < < ; ln() > 0 for >. Recall also that ln( 2 ) = ln + ln 2 Let us use implicit ifferentiation to fin ln(). w = ln(). We want to fin w. e w = (e w ) = () (e w ) w = w e w w = w = = e w (ln()) = 4

35 Lecture Fall 2006 Finally, what about (a )? There are two methos we can use: Metho : Write base e an use chain rule. Rewrite a as e ln(a). Then, ( ) a = e ln(a) = e ln(a) That looks like it might be tricky to ifferentiate. Let s work up to it: e e 3 = e an by the chain rule, = 3e 3 Remember, ln(a) is just a constant number not a variable! Therefore, e (ln a) (ln a) = (ln a)e or Recall that So now we know the value of M(a): (a ) = ln(a) a (a ) = M (a) a M(a) = ln(a). Even if we insist on starting with another base, like 0, the natural logarithm appears: 0 = (ln 0)0 The base e may seem strange at first. But, it comes up everywhere. After a while, you ll learn to appreciate just how natural it is. Metho 2: Logarithmic Differentiation. The iea is to fin f() by fining ln(f()) instea. Sometimes this approach is easier. Let u = f(). ( ) ln(u) u u ln(u) = = u u u Since u = f an = f, we can also write f (ln f) = or f = f(ln f) f 5

36 Lecture Fall 2006 Apply this to f() = a. ln f() = ln a = ln(f) = ln(a ) = ( ln(a)) = ln(a). (Remember, ln(a) is a constant, not a variable.) Hence, f (ln f) = ln(a) = = ln(a) = f = ln(a)f = a = (ln a)a f Eample. ( ) =? With variable ( moving ) eponents, you shoul use either base e or logarithmic ifferentiation. In this eample, we will use the latter. Therefore, f = ln f = ln ( ) (ln f) = (ln ) + = ln() + (ln f) = f f f = f(ln f) = (ln() + ) If you wante to solve this using the base e approach, you woul say f = e ln an ifferentiate it using the chain rule. It gets you the same answer, but requires a little more writing. ( ) k Eample 2. Use logs to evaluate lim +. k k Because the eponent k changes, it is better to fin the limit of the logarithm. [ ( )k ] lim ln + k k We know that [ ( )k ] ( ) ln + = k ln + k k ( ) This epression has two competing parts, which balance: k while ln + 0. k [ ( ) ] k ( ) ln ( + ) k ln( + h) ln + = k ln + = = (with h = ) k k h k k Net, because ln = 0 [ ( )k ] ln + ln( + h) ln() = k h 6

37 Lecture Fall 2006 Take the limit: h = k 0 as k, so that ln( + h) ln() lim = ln() = h 0 h = In all, ( ) k lim ln + =. k k ( ) k We have just foun that a k = ln[ + k ] as k. ( ) k If b k = + k, then b k = e a k e as k. In other wors, we have evaluate the limit we wante: ( ) k lim + = e k k Remark. We never figure out what the eact numerical value of e was. Now we can use this limit formula; k = 0 gives a pretty goo approimation to the actual value of e. Remark 2. Logs are use in all sciences an even in finance. Think about the stock market. If I say the market fell 50 points toay, you nee to know whether the market average before the rop was 300 points or 0, 000. In other wors, you care about the percent change, or the ratio of the change to the starting value: f (t) = ln(f(t)) f(t) t 7

38 Lecture Fall 2006 Lecture 7: Continuation an Eam Review Hyperbolic Sine an Cosine Hyperbolic sine (pronounce sinsh ): Hyperbolic cosine (pronounce cosh ): sinh() = e e e + e cosh() = 2 ( ) e e sinh() = = e ( e ) = cosh() 2 2 Likewise, cosh() = sinh() (Note that this is ifferent from cos().) Important ientity: cosh 2 () sinh 2 () = Proof: 2 cosh 2 () sinh 2 () = cosh 2 () sinh 2 () = ( ) 2 ( ) 2 e + e e e 2 2 ( ) ( ) 4 e 2 + 2e e + e 2 e e 2 = (2 + 2) = 4 4 Why are these functions calle hyperbolic? Let u = cosh() an v = sinh(), then u 2 v 2 = which is the equation of a hyperbola. Regular trig functions are circular functions. If u = cos() an v = sin(), then which is the equation of a circle. u 2 + v 2 =

39 Lecture Fall 2006 General Differentiation Formulas Eam Review (u + v) = u + v (cu) = cu (uv) = u v + uv (prouct rule) ( u ) u v uv = (quotient rule) v v 2 f(u()) = f (u()) u () (chain rule) You can remember the quotient rule by rewriting an applying the prouct rule an chain rule. ( u ) = (uv ) v Implicit ifferentiation Let s say you want to fin y from an equation like y 3 + 3y 2 = 8 Instea of solving for y an then taking its erivative, just take eample, of the whole thing. In this 3y 2 y + 6yy + 3y 2 = 0 (3y 2 + 6y)y = 3y 2 y = 3y 2 3y 2 + 6y Note that this formula for y involves both an y. Implicit ifferentiation can be very useful for taking the erivatives of inverse functions. For instance, Implicit ifferentiation yiels an y = sin sin y = (cos y)y = y = = cos y 2 2

40 Lecture Fall 2006 Specific ifferentiation formulas You will be responsible for knowing formulas for the erivatives an how to euce these formulas n from previous information:, sin, tan, sin, cos, tan, sec, e, ln. For eample, let s calculate sec : ( sin ) sec = = = tan sec cos cos 2 You may be aske to fin sin or cos, using the following information: Remember the efinition of the erivative: Tying up a loose en sin(h) lim = h 0 h lim cos(h) = 0 h 0 h f( + Δ) f() f() = lim Δ 0 Δ How to fin r, where r is a real (but not necessarily rational) number? All we have one so far is the case of rational numbers, using implicit ifferentiation. We can o this two ways: st metho: base e 2n metho: logarithmic ifferentiation = e ln r ( = e ln ) r = e r ln r = e r ln = e r ln r ln r (r ln ) = e ( r = r r ) = r r f (ln f) = f f = r ln f = r ln r (ln f) = ( ) f = f(ln f) = r r = r r 3

41 Lecture Fall 2006 Finally, in the first lecture I promise you that you learn to ifferentiate anything even something as complicate as tan e So let s o it! e uv = e uv (uv) = e uv (u v + uv ) Substituting, ( ( )) e tan = e tan tan