Calculus I (part 2): Introuction to Dierentiation (by Evan Dummit, 2016, v 250) Contents 2 Introuction to Dierentiation 1 21 Motivation: Rates an Tangent Lines 1 22 Formal Denition of the Derivative 3 23 Derivatives of Basic Functions 5 24 Calculating Derivatives 7 241 Rules for Computing Derivatives 8 242 Basic Examples of Derivatives 10 243 Aitional Examples of Derivatives 12 244 Logarithmic Dierentiation 14 25 Implicit Curves an Implicit Dierentiation 15 251 Implicit Relations an Implicit Functions 15 252 Implicit Dierentiation 17 26 Relate Rates 20 27 Linearization an Dierentials 21 271 Linearization, Approximation by the Tangent Line 21 272 Dierentials 23 2 Introuction to Dierentiation In this chapter, we evelop the erivative along with techniques for ierentiation We begin by ening the erivative as a limit an then use the enition to compute erivatives of simple functions We then prove the basic rules (such as the Prouct Rule an the Chain Rule) that will allow us to compute more complicate erivatives, an mention aitional time-saving techniques such as logarithmic ierentiation We then iscuss implicit relations an implicit functions (an how to compute their erivatives), an problems involving relate rates an the iea of the linearization of a function 21 Motivation: Rates an Tangent Lines We are often intereste in the rate at which processes occur, both in the physical sciences an the worl in general For example, we might want to know how fast a ball falls, how fast a chemical reaction occurs, how fast a population is growing Or we might be intereste in how fast our car is going, how fast money is accruing in our bank accounts, an how fast we are learning something We woul like a way to quantify this iea precisely The erivative of a function captures how fast that function is changing, at the point we measure it In other wors, the erivative measures the rate of change of a function 1
Think of looking at a car's speeometer: it measures how fast the car is moving at the specic instant the speeometer is rea One notion we alreay can talk about is average velocity: average velocity] =total istance] / total time] Example: A worl-class sprinter can run the 100m ash in 10 secons Over that time, the sprinter is traveling at an average spee of 10 meters per secon However, we are not asking for an average rate of change; we want to know the instantaneous rate of change, which is the velocity at a specic instant, rather than over a time interval One thing we might try is to calculate average rates of change over smaller an smaller intervals aroun our point of interest, to see if these numbers approach some value If the position function is given by f(t), an we want to look over the time interval t, t + t] where t is some very small time increment, then the average velocity over that time interval is average velocity] = en position] start position] f(t + t) f(t) = total time] t We woul like to examine this ierence quotient as we make the time interval t smaller an smaller We can o this precisely using limits, an this leas us to the formal enition of the erivative Another important use of the the erivative is in ning the equation for the line tangent to a curve at a point If f(x) is a nice function, then the tangent line to the graph y = f(x) at the point (a, f(a)) is the line passing through (a, f(a)) that just touches the graph at (a, f(a)), an, near that point of tangency, is the line which is closest to the graph of y = f(x) Here is a graph of y = 2x x 3 along with two tangent lines to the curve: To specify the equation of a line requires two pieces of information: the slope of the line an one point that the line passes through The tangent line to y = f(x) at x = a, by enition, passes through the point (a, f(a)), so in orer to etermine the tangent line, it is enough to n the slope of the tangent line One way we can try to compute the slope of the tangent line is by approximating the tangent line with secant lines passing through two points of the graph of y = f(x): through (a, f(a)) an another nearby point (a + h, f(a + h)), where h is very small The slope of the secant line through (a, f(a)) an (a + h, f(a + h)) is, by the slope formula, given by f(a + h) f(a) f(a + h) f(a) = (a + h) a h Then, if f behaves well enough, we shoul be able to n the slope of the tangent line by evaluating f(a + h) f(a) the slope of the secant line through (a, f(a)) an (a + h, f(a + h)), as h gets closer an h closer to 0 Again, using limits, we can formalize this notion, an (once again) the answer is given by the erivative 2
22 Formal Denition of the Derivative Denition: Provie the limit exists, we ene the erivative of f at x as f (x) = f x f(x + h) f(x) Alternate Denition: An alternate way of writing the enition of the erivative is f f(x) f(a) (a) x a x a Note that the limit in the alternate enition is the same as the limit in the enition above, upon making the change of variables h = x a A function whose erivative exists at x = a is calle ierentiable at x = a If a function is ierentiable for every value of x, we say it is everywhere ierentiable Most of the functions we will encounter are ierentiable everywhere that they are ene, although in general, functions o not nee to be ierentiable anywhere The erivative f (x) measures the instantaneous rate of change of f at x The erivative f (x) is also the slope of the tangent line to the graph of y = f(x) These are the most funamental interpretations of the erivative: the author of these notes calls them the calculus mantra Example: Use the enition of the erivative to compute f (x), for f(x) = x 2, an then n an equation for the tangent line to y = x 2 at x = 1 We have f (x + h) 2 x 2 (x) x 2 + 2xh + h 2 x 2 2xh + h 2 h 0 (2x + h) = 2x By this calculation, we know that the tangent line to y = x 2 at x = 1 has slope f (1) = 2 This tangent line passes through (1, 1), an thus by the point-slope formula has equation y 1 = 2(x 1), or y = 2x 1 Here is a graph of y = x 2 an its tangent line y = 2x 1 at x = 1: Non-Example: Show that the erivative of f(x) = x oes not exist at x = 0 If f(x) = x, then if we try to compute f h 0 (0), we have to evaluate the limit lim But this limit oes not exist because the two one-sie limits have unequal values, 1 an +1 Denition: The secon erivative of f, enote f (x) or 2 f, is the erivative of the erivative of f x2 h h 0 h In general we mostly care about the rst an secon erivatives of f(x), as they have the most meaning in physical applications But there is no reason to stop with just the secon erivative: in general, we ene the nth erivative of f, enote f (n) (x) or n f, to be the result of taking the erivative of f, n times xn 3
Example: Fin the secon an thir erivatives of f(x) = x 2 We alreay compute from the enition that f (x) = 2x So to compute the secon erivative of f, we just nee to n the erivative of 2x We have 2(x + h) 2x 2x + 2h 2x 2h 2x] x 2 = 2 h 0 So for f(x) = x 2, we see that f (x) = 2 (for every value of x) For the thir erivative, we nee to n the erivative of f (x) = 2: we obtain an so f (x) = 0 (for every value of x) 2 2 2] 0 = 0 x h 0 In the particular event that a function represents the position p(t) of an object moving along a one-imensional axis, with t representing time, the rst an secon erivatives have particularly concrete interpretations: Denition: If an object has position p(t) at time t, the rst erivative p (t) represents the velocity of the object, an the secon erivative p (t) represents the acceleration of the object Another way of phrasing these enitions is that velocity is the rate of change of position, an acceleration is the rate of change of velocity If the units of p(t) are istance, then the units of p (t) are istance per time, an the units of p (t) are istance per time square For example, if p(t) is measure in meters an t is measure in secons, then the units of p (t) are m /s (meters per secon) an the units of p (t) are m /s 2 (meters per secon square, or equivalently meters per secon per secon) These units come irectly out of the limit enition, because in the ierence quotient p p(t + h) p(t) (t), the numerator has units of istance while the enominator has units of time Likewise, in p (t) = p (t + h) p(t) lim, the numerator has units of istance per time, while the enominator still has units of time Remark: Derivatives past the secon erivative carry less obvious physical meaning, but the thir erivative p (t), which represents the rate of change of acceleration, is sometimes calle jerk Remarks on Notation: We will frequently write both f x an f (x) to enote the erivative of f with respect to x They mean the same thing, but we will typically use f (x) since it is easier to write, an use f x only when we nee to emphasize the ierence quotient nature of the erivative Another notation we will use for the erivative is f]: the symbol means take the erivative x of the thing that follows, with respect to We will generally use this notation when f is something complicate, to make it clearer what we are oing When f is a function of a single variable, the notation f always means the erivative of f with respect to that variable Taking aitional erivatives is enote by aing primes: thus f (x) is the thir erivative of f For functions involving more than one variable, the notation f is ambiguous, an shoul not be use For such functions, the notation f x is often use (in place of f ) to enote the erivative of f with respect to x On occasion, especially in physics, the notation f is use to enote the erivative of f with respect to a variable representing time (t), in contrast with the notation f use to enote the erivative of f with respect to a variable representing space (x) 4
Roughly speaking, a ierentiable function (one whose erivative exists everywhere) behaves nicely, at least in comparison to arbitrary functions We have another notion of niceness (namely, continuity), an it turns out that being ierentiable is a stronger conition than being continuous: Theorem (Dierentiable Implies Continuous): If f(x) is ierentiable at x = a, then f(x) is continuous at x = a This theorem says a ierentiable function is continuous The converse of this theorem is false: there exist functions which are continuous at a point but not ierentiable there, such as x at x = 0 Even more pathologically, there exist functions which are continuous everywhere but ierentiable nowhere Proof: By the multiplication rule for limits, we can write ] f(x) f(a) f(x) f(a) ] lim f(x) f(a)] (x a) lim x a x a x a x a x a (x a) = f (a) 0 = 0 x a where in the last step we use the alternate enition of the erivative f f(x) f(a) (a) x a x a Therefore, lim f(x) f(a)] = 0 Now since f(a) is a constant, we can pull it out of the limit an then x a rearrange the equality to rea lim f(x) = f(a) x a But this says precisely that f(x) is continuous at x = a, as we wante 23 Derivatives of Basic Functions These basic erivatives shoul be learne by heart: x c] = 0 x xn ] = n x n 1 x ex ] = e x x ln(x)] = 1 x sin(x)] x = cos(x) cos(x)] x = sin(x) Note 1: The formulas for the erivatives of sine an cosine require that the angle be measure in raians, not egrees This is the reason that we measure angles in raians: they are the most natural unit when oing calculus Note 2: In a similar way, the formula for the erivative of e x is surprisingly simple, while (as we will see) the erivative of the more general exponential function f(x) = a x is f (x) = a x ln(a), which has an unpleasant natural logarithm constant factor in it This is the reason we usually prefer to use the natural exponential e x in calculus, since its erivative is the simplest Likewise, among all the ierent logarithms, the one having the simplest erivative is the natural logarithm Here is the proof of each of these rules, from the enition of the erivative: The erivative of a constant is zero, since if f(x) = c for all x, then f c c 0 (x) = 0 5
If n is a positive integer, then for f(x) = x n, we have (by the alternate enition of the erivative) f x n a n (a) x a x a x a (x a)(x n 1 + x n 2 a + + x a n 2 + a n 1 ) x a x a ( x n 1 + x n 2 a + + x a n 2 + a n 1) = a n 1 + a n 1 + n terms + a n 1 = n a n 1 For other n (negative integers, rational numbers, general real numbers) the proof uses the same ieas, but is not much more enlightening For f(x) = e x, we have f (x) = e x+h e x lim = e x e h e x lim = e x (e h 1) lim = e x e h 1 lim = e x e h 1 because lim = 1, either by a calculation or by enition of the exponential For f(x) = ln(x), we have f ln(x + h) ln(x) (x) ( 1 ln 1 + h ) x ( 1 y ln 1 + 1 )] y x y = 1 x ln lim y x h 0 x/h ( ln 1 + h x = 1 x lim y y ln ( 1 + 1 ) y ] = 1 y x ln(e) = 1 x ) ( 1 + 1 y ( where we mae the change of variables y = x/h an use the fact that lim 1 + 1 ) y = e, either by a y y calculation or by enition of the number e For f(x) = sin(x), we have f (x) h 0 sin(x + h) sin(x) h cos(h) 1 sin(h) because lim = 0 an lim = 1 sin(x) cos(h) + cos(x) sin(h) sin(x) ( sin(x) cos(h) 1 + cos(x) sin(h) ) h 0 h h ] ] cos(h) 1 sin(h) = sin(x) lim + cos(x) lim = cos(x) 6 )
To obtain the sine limit, a geometry calculation shows that cos(x) < sin(x) < 1 for small positive x x Then apply the squeeze theorem as x 0 For the cosine limit, observe that cos(x) 1 = cos(x) 1 cos(x) + 1 x x cos(x) + 1 = sin 2 (x) x (cos(x) + 1) = sin(x) x sin(x), an then note that as x 0 the rst term goes to 1 while the secon goes to 0 cos(x) + 1 For f(x) = cos(x), we have f (x) h 0 cos(x + h) cos(x) h cos(h) 1 sin(h) again because lim = 0 an lim = 1 cos(x) cos(h) sin(x) sin(h) cos(x) ( cos(x) cos(h) 1 sin(x) sin(h) ) h 0 h h ] ] cos(h) 1 sin(h) = cos(x) lim sin(x) lim = sin(x) The erivatives of the other trigonometric functions, an the inverse trigonometric functions, often appear as well The most important are tan(x), sin 1 (x), an tan 1 (x) x tan(x)] = sec2 (x) sec(x)] = sec(x) tan(x) x x cot(x)] = csc2 (x) csc(x)] = csc(x) cot(x) x sin 1 (x) ] 1 = x 1 x 2 cos 1 (x) ] 1 = x 1 x 2 tan 1 (x) ] 1 = x 1 + x 2 sec 1 (x) ] 1 = x x x 2 1 csc 1 (x) ] 1 = x x x 2 1 cot 1 (x) ] = 1 x 1 + x 2 The formulas for tangent, secant, cosecant, an cotangent can be obtaine, with some eort, from the formal enition by reucing the calculations to sines an cosines However, it is much easier to obtain them using some of the rules for computing erivatives (which we have not iscusse yet) The erivatives of the inverse trigonometric functions can be compute using the enition of erivative, but it is complicate an rather tricky; we will avoi this an instea iscuss a general formula later for ning erivatives of inverse functions 24 Calculating Derivatives In this section, we rst iscuss all of the basic techniques for computing erivatives, an then give a number of examples an basic applications 7
241 Rules for Computing Derivatives In the list of rules below, f means f(x), f means f (x), an so on Using these rules, we can compute the erivatives of any elementary function (namely, any function which is a combination of sums, proucts, quotients, exponentials, logs, an trigonometric an inverse trigonometric functions of x) Sum/Dierence Rule: For any ierentiable f an g, we have x f + g] = f + g an x f g] = f g What this means is: if we have a sum (or ierence) of functions, we can compute the erivative of the sum (or ierence) just by ierentiating each term one at a time, an then aing (or subtracting) the results Proof: We have f(x + h) + g(x + h)] f(x) + g(x)] f + g] x h 0 h f(x + h) f(x)] + g(x + h) g(x)] f(x + h) f(x)] g(x + h) g(x)] + lim = f (x) + g (x) The proof for f g is the same, except with a minus sign It might seem, base on the rules for sums an ierences, that the erivative of a prouct of two functions woul be the prouct of their erivatives However, this is not true! For example, x 5 = x 3 x 2, but the erivative of x 5 is 5x 4 while the prouct of the erivatives of x 3 an x 2 is 3x 2 2x = 6x 3 The correct rule turns out to be slightly more complicate Prouct Rule: For any ierentiable functions f an g, we have x f g] = f g + g f In the case where g is the constant function c, we obtain the Constant Multiple Rule Proof: We have x c f] = c f f(x + h)g(x + h) f(x)g(x) f(x)g(x)] x h 0 h f(x + h)g(x + h) f(x + h)g(x)] + f(x + h)g(x) f(x)g(x)] ] ] ] ] g(x + h) g(x) f(x + h) lim + lim 0 h g(x) f(x + h) f(x) lim 0 h = f(x)g (x) + g(x)f (x) where in the last step, we use the fact that lim h 0 f(x + h) = f(x) because f(x) is ierentiable an therefore continuous There is also a rule for computing erivatives of quotients Like with the Prouct Rule, the erivative of a quotient is not the corresponing quotient of erivatives, but rather something slightly more complicate 8
Quotient Rule: For any ierentiable functions f an g with g 0, we have x ] f = f g g f g g 2 In the Quotient Rule, it is very important not to mix up the orer of the two terms in the numerator (f g has a positive sign, while g f has a negative sign) Getting them backwars will yiel a result that is o by a factor of 1 Proof: First we show that x ] 1 = g g g : 2 1 ] x g(x) h 0 1 g(x + h) 1 g(x) h ( ) g(x) g(x + h) g(x + h) g(x) g(x) g(x + h) 1 g(x) g(x + h) ] ] g(x) g(x + h) 1 lim h 0 g(x) g(x + h) = g 1 (x) g(x) 2 where in the last step, we use the enition of the erivative an fact that lim g(x + h) = g(x) because h 0 g(x) is ierentiable an therefore continuous The general result for f g follows by writing f g = f 1 an applying the Prouct Rule: g x ] f = f 1 ] g x g = x f] 1 g + f x = f 1 g + f g g 2 = f g g 2 g f g 2 ] 1 g = f g g f g 2 The rules above allow us to ierentiate any combination of sums, ierences, proucts, an quotients However, it is not possible to escribe a function like h(x) = sin(cos(x)) using only those operations an functions whose erivatives we know The missing ingreient is function composition: notice that h(x) = f(g(x)) where f(x) = sin(x) an g(x) = cos(x) We will now give a rule for ning the erivative of a composition chain like h(x) = f(g(x)) Chain Rule: For any ierentiable functions f an g, we have x f(g(x))] = f (g(x)) g (x) Another way of writing the Chain Rule is z x = z y y x, where z is a function of y an y is a function of x The translation between this formulation an the explicit one above is y = g(x) an z = f(y) = f(g(x)) A natural way to interpret this formulation of the Chain Rule is using rates of change: for example, if z changes twice as fast as y an y changes four times as fast as x, then z shoul be changing 8 times as fast as x 9
Proof 1 : By the alternate enition of the erivative, f(g(x)) f(g(a)) f(g(a))] x x a x a f(g(x)) f(g(a)) x a g(x) g(a) f(g(x)) f(g(a)) = x a lim y b g(x) g(a) ] f(y) f(b) y b = f (b) g (a) = f (g(a)) g (a) g(x) g(a) x a ] ] g(x) g(a) lim x a x a ] g(x) g(a) x a lim x a where to get from the thir equation to the fourth, we substitute y = g(x) an b = g(a) in the rst limit an use the fact that y b as x a because g(x) is ierentiable an therefore continuous There are a few useful corollaries of the Chain Rule which are worth remembering on their own: When f(x) is the natural logarithm, we obtain the Log Rule g ln(g)] = x g When f(x) is the base-e exponential we obtain the Exponential Rule e g(x)] = e g(x) g (x) x In particular, because a x = e x ln(a), if we set g(x) = x ln(a) in the rule above, we obtain the general-base exponential erivative x ax ] = a x ln(a) Finally, we will give a rule for computing the erivative of an inverse function: Inverse Rule: If f is a ierentiable function with inverse f 1, then x f 1 1 (x)] = f (f 1 (x)) Proof: By the alternate enition of erivative, we have f 1 (x) ] f 1 (x) f 1 (a) (a) x x a x a Now set b = f 1 (a) an make the substitution y = f 1 (x), so that a = f(b) an x = f(y) We obtain f 1 (x) ] y b (a) x y b f(y) f(b) = 1 = 1 f(y) f(b) f lim (b) = 1, as esire f (f 1 (a)) y b y b 242 Basic Examples of Derivatives The best way to become comfortable computing erivatives is to work examples Here are some simpler ones: Example: Dierentiate f(x) = x with respect to x Note that x = x 1/2, so applying the Power Rule yiels f (x) = 1 2 x 1/2 Example: Dierentiate g(x) = 3x 10 + 3x 1 x 1/2 with respect to x 1 This proof has a minor issue, which is that g(x) g(a) might be zero even for x very close to a; an example occurs for a = 0 with the function g(x) = x 2 f(g(x)) f(g(a)) cos(1/x) To x this issue one must analyze the function more carefully; the precise etails g(x) g(a) are not especially enlightening an so we omit them 10
Here, we can just ierentiate term-by-term (secretly applying the Sum an Dierence Rules) using the Power Rule to get g (x) = 30x 9 3x 2 1 2 x 1/2 Example: For h(y) = e y ln(y), n h (y) The Prouct Rule gives h (y) = e y y ln(y)] + y ey ] ln(y) = e y 1 y + ey ln(y) Example: Dierentiate q(t) = et + 3 t 2 with respect to t The Quotient Rule gives q (t) = t et + 3] (t 2 ) t t2 ] (e t + 3) (t 2 ) 2 = et (t 2 ) 2t (e t + 3) (t 2 ) 2 This can be simplie by cancelling a factor of t, giving q (t) = t et 2(e t + 3) t 3 Example: Dierentiate f(x) = tan(x) with respect to x If we write f(x) = sin(x)/ cos(x), we can apply the Quotient Rule to see f (x) = = x sin(x)] (cos(x)) x cos(x)] (sin(x)) cos 2 (x) cos(x) cos(x) ( sin(x)) sin(x) cos 2 (x) = cos2 (x) + sin 2 (x) cos 2 (x) = sec 2 (x) = 1 cos 2 (x) The erivatives of secant, cosecant, an cotangent can be compute the same way Example: Fin the erivative of h(x) = (x 2 + e x ) 15 Here, we can apply the Chain Rule, with f(x) = x 15 an g(x) = x 2 + e x We obtain h (x) = f (g(x)) g (x) = 15(x 2 + e x ) 14 (2x + e x ) Example: Dierentiate k(x) = x 3 + 1 Here, we can apply the Chain Rule, with f(x) = x = x 1/2 an g(x) = x 3 + 1 We obtain k (x) = f (g(x)) g (x) = 1 2 (x3 + 1) 1/2 3x 2 Example: For g(x) = tan 1 (x), n g (x) We use the Inverse Rule: notice that g(x) = f 1 (x) for f(x) = tan(x) Since f (x) = sec 2 (x), we get g (x) = 1 f (f 1 (x)) = 1 1 + tan 2 (tan 1 (x)) = 1 1 + x 2 The erivatives of the other inverse trigonometric functions can be compute the same way Using these ierentiation techniques, we can quickly solve the kins of problems that motivate our enition of the erivative without having to resort to cumbersome limit calculations: Example: Fin an equation for the tangent line to y = x 2 + 3 x at x = 1 11
We have y(1) = 4, so the tangent line passes through (1, 4) The erivative of this function is y = 2x 3 x 2, so y (1) = 1 Therefore, the tangent line has slope 1 an passes through the point (1, 4), so its equation is y 4 = 1(x 1) (Note that we coul also write the equation in a variety of other ways, such as y = 5 x or x + y = 5) Example: If at time t secons, a particle has position p(t) = t sin(πt) meters, how fast is it moving at t = 2s an how fast is it accelerating at t = 3s? The velocity is given by the erivative of position, so v(t) = p (t) = 1 π cos(πt), so at t = 2s, we see that v(2) = 1 π cos(2π) = 1 π The units on the velocity are meters per secon, an since 1 π is negative an spee is positive, at time t = 2s we see that the particle is moving at (π 1) m /s Acceleration is the erivative of velocity, an the units on acceleration are meters per secon square, m/s 2, so a(t) = v (t) = π 2 sin(πt), so at t = 3s, we see that a(3) = 0 m /s 2 (This says that the particle is not accelerating either forwars or backwars at time t = 3s) Example: Fin an equation for the tangent line to y = x 1/3 at x = 1, at x = 1 8, at x = 1, an explain what 27 happens at x = 0 The erivative is y = 1 3 x 2/3 We have y(1) = 1 so the rst line passes through (1, 1) Its slope is y (1) = 1, so its equation is 3 y 1 = 1 3 (x 1), or y = 1 3 x + 2 3 We have y( 1 8 ) = 1 2, an y ( 1 8 ) = 4 3, so the secon line's equation is y 1 2 = 4 3 (x 1 8 ), or y = 4 3 x + 1 3 We have y( 1 27 ) = 1 3, an y ( 1 27 ) = 3, so the thir line's equation is y 1 3 = 3(x 1 27 ), or y = 3x + 2 9 At the origin (0, 0), we see that the erivative y is unene: if we take the limit as x 0 from either sie, we see that y + (We will often abuse the terminology slightly an say that the erivative is innite in a case like this) Uner the usual interpretation of a line with slope as a vertical line, we guess that the tangent line to this curve is vertical at the origin an thus has equation x = 0 Inee, this is borne out by the graph of the curve an the three tangent lines we foun above the tangent line at the origin oes inee look like a vertical line: 10 10 10 10 05 05 05 05 05 10 15 05 10 15 05 10 15 05 10 15 05 05 05 05 243 Aitional Examples of Derivatives The examples we have given so far involve essentially a single application of one of the ierentiation rules However, suciently complicate functions will often require use of several of the rules in tanem Although it may seem icult, the proceure is very algorithmic: simply ecie which which rule can be applie rst Then write own the result of applying that rule to the given expression, an then evaluate any remaining erivatives in the same way ( Example: Dierentiate f(x) = ln 1 + e ) x 12
First, observe that we can apply the Chain Rule to see that f x (x) = (1 + e x ) 1 + e x Now we are left with the simpler task of evaluating the erivative in the numerator We can o this with another application of the Chain Rule, yieling the en result f (x) = e x 1 2 x 1/2 1 + e x Example: Dierentiate f(x) = x3 + 2 x sin(x) We start by applying the Quotient Rule, giving f (x) = x x3 + 2 x] sin(x) x sin(x)] (x3 + 2 x) (sin(x)) 2 Evaluating the leftover erivatives yiels f (x) = (3x2 + x 1/2 ) sin(x) cos(x) (x 3 + 2 x) (sin(x)) 2 Example: Dierentiate h(r) = r 3 e r ln(r) Here, we have a prouct of 3 terms We can apply the Prouct Rule to this expression by grouping two of the terms together: h(r) = r 3 e r] ln(r) We obtain h (r) = r 3 e r] ln(r) + r 3 e r r r ln(r)] = 3r 2 e r + r 3 e r] ln(r) + r 3 e r 1 r where we use the Prouct Rule again to ierentiate r 3 e r Example: Dierentiate f(x) = sin(sin(sin(sin(x)))) This function seems intimiating, but it is built entirely through function composition, so we can n its erivative using the Chain Rule by working from the outsie to the insie To start, note that f(x) has the form sin( ), where is some other function (namely, sin(sin(sin(x)))) Therefore, f (x) = cos( ) ] by the Chain Rule x We have now reuce the problem to ning the erivative of the simpler expression = sin(sin(sin(x))) Repeately applying the Chain Rule in this manner yiels f (x) = cos(sin(sin(sin(x)))) x sin(sin(sin(x)))] = cos(sin(sin(sin(x)))) cos(sin(sin(x)) Example: Dierentiate f(x) = x tan(x 3 + e 2x2 cos(x) ) x sin(sin(x))] = cos(sin(sin(sin(x)))) cos(sin(sin(x)) cos(sin(x)) x sin(x)] = cos(sin(sin(sin(x)))) cos(sin(sin(x)) cos(sin(x)) cos(x) By applying the Prouct Rule an Chain Rule repeately, we obtain f (x) = x x] tan(x3 + e 2x 2 cos(x) ] ) + x tan(x 3 + e 2x2 cos(x) ) x = 1 tan(x 3 + e 2x2 cos(x) ) + x sec 2 (x 3 + e 2x2 cos(x) ) x ] 3 + e 2x2 cos(x) x = 1 tan(x 3 + e 2x2 cos(x) ) + x sec 2 (x 3 + e 2x2 cos(x) ) 3x 2 + e 2x2 cos(x) 2x 2 cos(x) ]] x = 1 tan(x 3 + e 2x2 cos(x) ) + x sec 2 (x 3 + e 2x2 cos(x) ) 3x 2 + e 2x2 cos(x) 4x cos(x) 2x 2 sin(x) ]] 13
244 Logarithmic Dierentiation Evaluating erivatives of large proucts, quotients, an complicate powers can be achieve using the rules we have establishe, but the results can be very cumbersome There is a useful trick to simplify such ierentiation problems, calle logarithmic ierentiation: the basic iea of the proceure is to ierentiate ln(f) rather than f itself If ln(f) can be simplie using properties of logarithms, the resulting computations can be substantially easier We will illustrate the iea with a few examples Example: Dierentiate f(x) = (x 2) 3 (x 2 + 1) 7 We coul just use the Prouct an Chain Rules, giving f (x) = 3(x 2) 2 (x 2 + 1) 7 + (x 2) 3 7(x + 1) 6 2x Let us now solve the problem a ierent way: if we rst take logarithms of both sies, we obtain ln(f) = ln (x 2) 3 (x 2 + 1) 7] = 3 ln(x 2) + 7 ln(x 2 + 1) Dierentiating both sies then yiels f f = 3 1 x 2 + 7 2x x 2 + 1 ] Thus, f = f 3 = (x 2) 3 (x 2 + 1) 7 3 1 x 2 + 7 2x x 2 + 1 1 x 2 + 7 ] 2x x 2 + 1 If we expan out the prouct an cancel enominator terms, we get f = 3(x 2) 3 (x 2 +1) 7 +7(x 2) 3 (x+ 1) 6 2x, which is the same expression we obtaine above (Of course, this shoul not be surprising!) The logarithmic ierentiation metho saves the most time with complicate proucts, quotients, or powers Example: Dierentiate f(x) = (x 3)7 (x 2 + 1) 8 (x 3 + 2) 4 In principle, we coul use the Quotient, Prouct, an Chain Rules to solve this problem, but the calculations woul be rather lengthy (x 3) 7 (x 2 + 1) 8 ] Instea, let us write ln(f) = ln (x 3 + 2) 4 = 7 ln(x 3) + 8 ln(x 2 + 1) 4 ln(x 3 + 2) Dierentiating both sies yiels f 7 Thus, f = f x 3 + 8 Example: Dierentiate f(x) = f = 7 1 x 3 + 8 2x x 2 + 1 4 3x 2 x 3 + 2 x3 + 3 5 (x 3 1) 2 (x 4 + x + 1) 100 2x x 2 + 1 4 3x 2 x 3 + 2 ] = (x 3)7 (x 2 + 1) 8 (x 3 + 2) 4 We have ln(f) = 1 2 ln(x3 + 3) + 2 5 ln(x3 1) 100 ln(x 4 + x + 1) Dierentiating yiels f Thus, f = f = 1 2 x3 + 3 5 (x 3 1) 2 (x 4 + x + 1) 100 Example: Dierentiate f(x) = (x + sin(2x)) tan(x) We write ln(f) = tan(x) ln x + sin(2x)] 3x 2 x 3 + 3 + 2 3 3x 2 x 3 1 100 4x 3 x 4 + x + 1 1 2 3x 2 x 3 + 3 + 2 3 7 x 3 + 8 3x 2 x 3 1 100 4x 3 x 4 + x + 1 Dierentiating yiels f f = sec2 (x) ln x + sin(2x)] + tan(x) 1 + 2 cos(2x) x + sin(2x) Thus, f = (x + sin(2x)) tan(x) sec 2 (x) ln x + sin(2x)] + tan(x) 1 + 2 cos(2x) ] x + sin(2x) ] 2x x 2 + 1 4 3x 2 ] x 3 + 2 14
More generally, we can use logarithmic ierentiation to n the erivative of a general exponential of the form f(x) g(x) : the result is ] f(x) g(x) = g (x) ln f(x) + g(x) f ] (x) f(x) g(x) x f(x) It is not worth memorizing this rule (which is why it is not boxe) Instea, it is much more useful to remember how to compute such erivatives using logarithmic ierentiation 25 Implicit Curves an Implicit Dierentiation With the ierentiation rules, we can now compute the erivative of any function y = f(x) ene explicitly as a function of x, at least as long as it involves polynomials, powers, exponentials, logarithms, or (inverse) trigonometric functions (whose erivatives we know) Sometimes, however, we are also intereste in implicit relations, which relate the values of y an x not via an explicit function y = f(x), but as some equation of the form R(x, y) = 0, where R is an arbitrary function of the two variables x an y Any equation relating x an y can (by moving all the terms to one sie) be written in the form R(x, y) = 0 So an implicit relation involving x an y is the same thing as some equation that x an y satisfy 251 Implicit Relations an Implicit Functions Denition: An implicit relation between two variables x an y is an equation of the form R(x, y) = 0, where R is an arbitrary function of the two variables x an y We allow ourselves to rearrange implicit relations, to put things on either sie of the equality Examples: x + y = 1, x 2 + y 2 = 1, x + y 2 e y = 1, an x y + y cos(x) = tan(x y) are four implicit relations between x an y Sometimes, it is possible to turn an implicit relation into an explicit one by rearranging it Other times, this is not possible Example: Rearranging x + y = 1 yiels the explicit formula y = 1 x for y in terms of x Example: Rearranging x 2 +y 2 = 1 yiels the explicit formula y = ± 1 x 2 Note that the plus-or-minus is necessary, because both y = 1 x 2 an y = 1 x 2 satisfy x 2 + y 2 = 1 Example: There is no obvious way to solve x+y 2 e y = 1 to write y explicitly as a function of x However, it is possible to write x as a function of y: x = 1 y 2 e y Example: There is no obvious way to solve x y + y cos(x) = tan(x y) explicitly for either one of the variables in terms of the other Like with explicit functions y = f(x), we can prouce a graph of all the points (x, y) satisfying an implicit relation Note that actually proucing the graph of an implicit relation, though, is trickier than proucing the graph of an explicit function y = f(x): more is require than just picking a bunch of values of x an evaluating the function Usually, the graph of a single implicit relation will be a curve or a collection of curves (possibly intersecting each other) in the plane Here are graphs of x + y = 1, x 2 + y 2 = 1, x + y 2 e y = 1, an x y + y cos(x) = tan(x y), respectively: 15
As can be seen from the plot of x y + y cos(x) = tan(x y), implicit relations can have very complicate graphs! Motivating Example: Consier the implicit relation x 2 + y 2 = 1, whose graph is the unit circle To escribe the graph of the unit circle explicitly we have to glue together the graphs of the two functions y = 1 x 2 an y = 1 x 2 for 1 x 1 But if we allow ourselves an implicit relation for y, then we can escribe the unit circle with the far simpler single equation x 2 + y 2 = 1 In the case of x 2 + y 2 = 1, we can solve the implicit relation for y However, for more complicate implicit relations, it may not be possible to solve for y explicitly as a function of x; for example, it is not possible to solve y +sin(y) = x+e x for either x or y as an elementary function of the other Even if it is possible to solve for y, the result might be very complicate An it is also possible to lose information, if we are not extremely careful about solving the relation for y For example, x = sin(y) an y = sin 1 (x) appear to be equivalent, but in fact they aren't: (x, y) = (0, 2π) satises the rst equation but not the secon one, because the (principal) arcsine function sin 1 only has range π 2, π ] 2 Remark (relation versus function) : We often want to speak of implicit relations between x an y as implicitly ening y as a function of x Technically speaking, this is often troublesome: in any given implicit relation, there may be more than one value of y associate to a particular value of x, in which case y is not actually a function of x If we want to insist on trying to make y an actual function of x, we woul have to specify which particular value of y we want whenever there is a choice 16
For example, consier the implicit relation x 2 = y 2 If we solve for y in terms of x then we obtain y = ±x For each value of x (except for x = 0) the relation gives two possible values of y If we wante to solve the implicit relation x 2 = y 2 to obtain y as a function of x, we woul nee to specify which value of y we want for each possible value of x There are many ways of oing this: some of the obvious ones are y = x, or y = x, or y = x But we coul also o something like taking y = x, for 3 x 3, an taking y = x for the other possible x From here onwars, we will abuse terminology an not worry about whether the things we want to call implicit functions shoul really be name functions or not 252 Implicit Dierentiation One very natural question we might have is: if we have a point (x, y) on an implicit curve, what happens to y if we change x by a small amount? If we ha a curve given explicitly, changing x by a small amount x will cause y to change by a small amount y, an we will have an approximate equality y x y x = y So the question becomes: what is the right way to think about an compute erivatives implicitly? Here is the answer: To calculate the erivative y, in a situation where y an x satisfy an implicit relation R(x, y) = 0, we think of y as y(x), an implicit function of x, an then ierentiate both sies of the relation R(x, y) = 0 via the Chain Rule This will give an equation which can be solve for y as a function of x an y Note that the interpretation of y is still exactly the same: y still gives the slope of the tangent line to the curve Thus, when y = 0, the tangent line is horizontal, an when y blows up to ±, the tangent line is vertical One other type of unusual behavior can happen with implicit relations: if the graph of the curve R(x, y) = 0 crosses itself, then the value of y will usually be an ineterminate fraction of the form 0 0 Furthermore, if we want to compute the secon erivative y in terms of x an y, then we can simply ierentiate y (using the Chain Rule), an then substitute in for any occurrences of y in the resulting expression to write y as a function of x an y Repeately ierentiating in this manner allows calculation of any higher erivative, although the results usually get complicate very quickly Example: For the implicit function y ene by x 2 + y 2 = ( 1, n) y using implicit ierentiation, an then 3 n the slope of the tangent line to its graph at the point 5, 4 5 To n y, we think of y as a function of x an ierentiate both sies of x 2 + y(x)] 2 = 1 with the Chain Rule This gives 2x + 2 y(x)] 1 y (x) = 0, or, more compactly, 2x + 2y y = 0 Solving for y gives 2y y = 2x, so that y = 2x 2y = x y ( 3 5, 4 5 ) 2 = 1, is then y = 3/5 4/5 = 3 4 The slope of the tangent line at ( 5) 3 2 ( 4 + 5 ), which we can verify actually oes lie on the curve since Example (repeate): For the implicit function y ene by x 2 +y 2 = 1, ( try to) n y by solving for y explicitly, 3 an then n the slope of the tangent line to its graph at the point 5, 4 5 17
Solving x 2 + y 2 = 1 for y gives y = ± 1 x 2 = ±(1 x 2 ) 1/2 Note that the ± is very much necessary because there are two possible values of the square root, positive an negative If we blinly take the erivative of the expression y = ±(1 x 2 ) 1/2 with the Chain Rule, we obtain y = ± 1 2 (1 x2 ) 1/2 ( 2x) Now there is alreay an issue: which sign o we want? The only plausible answer is that it must epen on the value of y we are intereste in For y = 4/5, in particular, we probably want the negative sign Proceeing with this assumption, plugging in x = 3 5 to the expression y = 1 2 (1 x2 ) 1/2 ( 2x) gives y = 1 ( ) ] 2 3 2 1 1/2 ( 2 3 5 5 ), which after some arithmetic eventually oes simplify to the answer y = 3 4 we foun above Now compare our general formula from the explicit solution to the problem, to the general formula to the implicit solution: the formula y = ± 1 2 (1 x2 ) 1/2 ( 2x) is far more complicate (not to mention the ambiguity about the sign) than the formula y = x y In this case, we can see that the implicit formula is much better (This is almost always the case for implicit ierentiation problems!) Example: Fin an equation for the tangent line to the curve x 2 + xy + y 4 = 3 at the point (x, y) = (1, 1) We have a point the tangent line passes through, namely (1, 1), so all we nee is the line's slope, which is the erivative y To n y, we think of y as an implicit function y(x), an write the equation as x 2 +x y(x)]+y(x)] 4 = 3 Now we ierentiate both sies using the Chain Rule an the Prouct Rule, to get 2x + (1 y(x) + x y (x)) + 4 y(x)] 3 y (x) = 0, or, more compactly, 2x + y + xy + 4y 3 y = 0 Rearranging gives (2x + y) + (x + 4y 3 )y = 0, or y = 2x + y x + 4y 3 Plugging in the point (1, 1) gives y (1, 1) = 3 5 Thus the tangent line has slope 3 5 an passes through (1, 1), so an equation for the tangent line is y 1 = 3 (x 1) 5 Here is a graph of the curve together with this tangent line: 18
Example: Fin y an y for the curve y 2 = x 3 + x 2 Examine what happens to y at the point ( 1, 0) an at the point (0, 0) Dierentiating both sies of y(x)] 2 = x 3 + x 2 via the Chain Rule gives 2 y(x)] y (x) = 3x 2 + 2x Solving for y gives y = 3x2 + 2x 2y To n y, we ierentiate the expression for y with respect to x: this yiels y = x Substituting in for y gives y = 3x 2 + 2x ] (2y) (3x 2 + 2x) x 2y] (2y) 2 = (6x + 2) (2y) (3x2 + 2x) (2y ) (2y) 2 (6x + 2) (2y) (3x 2 + 2x) 3x2 + 2x y (2y) 2 = (6x + 2) (2y) y (3x2 + 2x) 2 4y 3 At ( 1, 0), the formula gives y = 1, which is the 'innite' type of unene, so here the tangent line is 0 vertical, as is conrme by the graph At (0, 0), the formula gives y = 0, which is the 'ineterminate' type of unene From the graph, it 0 appears that the curve crosses itself at the origin, an it even looks (in some sense) like there are actually two tangent lines at the origin In fact, this suspicion is accurate, per the following calculation: if we solve for y, we obtain y = x x + 1 or y = x x + 1 For x 0, setting y = x x + 1 yiels y = 3x2 + 2x 2y limit as x 0 yiels y = 1 = 3x2 + 2x 2(x x + 1) = 3x + 2 2, an now taking the x + 1 In a similar way, setting y = x x + 1 an then taking the limit as x 0 yiels y = 1 Thus, the two 'tangent lines' at the origin have slopes 1 an 1, meaning that they are y = x an y = x Remark: Curves with the general form y 2 = x 3 + a x 2 + b x + c are calle elliptic curves (The curve in this example is calle a singular elliptic curve, because the graph crosses itself) Elliptic curves have a rather surprisingly wie variety of applications: in particular, they are use extensively in cryptography (the secure transmission of information) an are the basis for some algorithms for factorization of large integers 19
26 Relate Rates If we have a relation between two quantities, then by ierentiating that relation an applying the Chain Rule we obtain a relationship between the erivatives of these quantities Therefore, if we know the rate that one quantity is changing, we can n the rate at which the other is changing These types of problems are calle relate rates problems, as they involve analyzing the rate of change of one quantity using information about relate quantities To solve a relate rates problem, follow these steps: First, if not alreay given, choose variable names an translate the given information into mathematical statements Ientify all relations between the variables Draw a picture, if the relations are not immeiately clear Next, ierentiate each relation using the Chain Rule Finally, solve for the esire quantity using the given information an the information obtaine from the erivatives Example: A spherical balloon is inate such that its volume increases by 40π cm 3 per secon Fin the rate at which the balloon's raius is increasing (i) when the raius is 2 cm, an (ii) when the raius is 5 cm The volume V of the balloon is given by V = 4 3 πr3 (in cm 3 ), where r is the raius of the balloon (in cm) We are given that V t = 40πcm3 /s an want to n r t Dierentiating the relation V = 4 3 πr3 via the Chain Rule gives V t Then, solving for r r gives t t = V/t 4πr information When r = 2 we see r t = When r = 5 we see r t = = 4πr2 r t 2 Now we can plug in the appropriate values to n the esire 40π cm 3 /s 4π 2 2 cm 2 = 5 2 cm /s So when r = 2, the raius is increasing at 25 cm /s 40π cm 3 /s 4π 5 2 cm 2 = 2 5 cm /s So when r = 5, the raius is increasing at 04 cm /s Note that we obtain the correct units ( cm /s, istance per time) for the rate of change r t Example: Two ships leave a port, one traveling north an the other traveling east After 4 hours, the northmoving ship is moving at 40 km /h an is 120 km from the port, an the east-moving ship is moving at 30 km /h an is 90 km from the port Fin the rate at which the istance between the ships is changing after 4 hours Let the istance of the east-traveling ship from the port be x(t) an the istance of the north-traveling ship from the port be y(t), where t is measure in hours Then by the Pythagorean Theorem, the istance (t) between the ships satises (t) 2 = x(t) 2 + y(t) 2 We want to n (t) The given information says that x(4) = 120 km, x (4) = 40 km /h, y(4) = 90 km, an y (4) = 30 km /h Plugging in t = 4 gives (4) 2 = x(4) 2 + y(4) 2 = (120 km) 2 + (90 km) 2 = 150 2 km 2, so (4) = 150 km Now, ierentiating the relation (t) 2 = x(t) 2 + y(t) 2 via the Chain Rule gives 2 (t) (t) = 2 x(t) x (t) + 2 y(t) y (t) Upon solving for the esire quantity (t), we see (t) = x(t) x (t) + y(t) y (t) (t) If we then take t = 4 an plug in all of the known values, we get (4) = x(t) x (t) + y(t) y (t) = (120 km)(40 km /h) + (90 km)(30 km /h) 150 km = 7500km2 /h 150 km = 50 km /h 20 (t)
Example: A large rectangular box has a length of 6 m, a with of 4 m, an a height of 3 m The length is changing at a rate of +1 m /s, the with is changing at a rate of 2 m /s, an the height is changing at a rate of +1 m /s Fin the rates at which the volume an total surface area of the crate are changing Let the length be l(t), the with be w(t), an the height be h(t) We are given l = 6 m, w = 4 m, h = 3 m, l = +1 m /s, w = 2 m /s, an h = +1 m /s, all at time t = 0 The volume is given by V = l w h an the surface area is given by A = 2lw + 2lh + 2wh Dierentiating each expression yiels V t = t (lw) h] = h lw] h + (lw) t t = an ( l t w + l w ) h + (lw) h t t = l w h wh + l h + lw t t t A t = l 2lw + 2lh + 2wh] = 2 t t w + l w t + l t h + l h t + w t h + w h ] t Now we just nee to plug in all of the given ata We obtain an V = (1 m /s)(4 m)(3 m) + (6 m)( 2 m /s)(3 m) + (6 m)(4 m)(1 m /s) = 0 m3 /s A = 2 (1 m /s)(4 m) + (6 m)( 2 m /s) + (+1 m /s)(3 m) + (6 m)(1 m /s) + ( 2 m /s)(3 m) + (4 m)(+1 m /s)] = 2 4 m2 /s 12 m2 /s + 3 m2 /s + 6 m2 /s 6 m2 /s + 4 m2 /s] = 2 m 2 /s So the volume of the box is not changing, an the surface area is increasing at 2 m2 /s 27 Linearization an Dierentials Another use of erivatives is approximating complicate functions by simpler ones We are particularly intereste in oing this in physics an economics, because it is easier to unerstan the behavior of simple functions This is especially important in the case where the relation between two variables can only be written implicitly, an we woul like to be able to say something about what happens to one variable if the other changes 271 Linearization, Approximation by the Tangent Line The key insight is: the tangent line to the graph of a function is a goo approximation to the function near the point of tangency Here is a justication of this iea: The enition of the erivative says that f f(x 0 + x) f(x 0 ) (x 0 ) (Note that is the Greek x 0 x letter elta, an is intene to inicate a small change in the variable following it) Intuitively, the limit is saying that that for small values of x, the quotient f(x + x) f(x 0) shoul x be close to the value f (x 0 ), which we write as f(x + x) f(x 0) f (x 0 ) (The symbol means x approximately equal to) Clearing the enominator an rearranging yiels small Equivalently, if we write x = x 0 + x, this says (ie, when x is close to x 0 ) f(x 0 + x) f(x 0 ) + f (x 0 ) x, where x is f(x) f(x 0 ) + f (x 0 ) (x x 0 ), where x x 0 is small 21
Note how similar this expression is to the equation of the tangent line to y = f(x) at x = x 0, which is y = f(x 0 ) + f (x 0 ) (x x 0 ) Denition: If f(x) is ierentiable at x = x 0, then the linearization of f(x) at x = x 0 is ene to be the linear function L(x) = f(x 0 ) + f (x 0 ) (x x 0 ) Example: Fin the linearization of the function (1 + x) 3 at x = 0 We have f(x) = (1 + x) 3 an x 0 = 0, an also f (x) = 3(1 + x) 2 by the Chain Rule Applying the formula shows L(x) = f(0) + f (0) (x 0) = 1 + 3x The reason we are intereste in the linearization is because it is the best linear approximation to a ierentiable function: Theorem (Linear Approximation): If f(x) is ierentiable at x = x 0, then the linearization of f(x) near x = x 0, ene by L(x) = f(x 0 ) + f f(x) L(x) (x 0 ) (x x 0 ), is the only linear function satisfying lim = 0 x x 0 x x 0 Equivalently, this result says that the tangent line y = L(x) is the only function with the property that, when we approximate f(x) with L(x), the resulting error near x = x 0 is small when compare to x x 0 Proof: Suppose f(x) is ierentiable at x = x 0 f(x) L(x) f(x) f(x 0 ) f (x 0 ) (x x 0 ) f(x) f(x 0 ) We compute lim f (x 0 ) = 0 x x 0 x x 0 x x 0 x x 0 x x 0 x x 0 by the enition of the erivative, so the linearization oes have the esire property f(x) g(x) Now suppose that g(x) = a + b(x x 0 ) is another linear function with lim = 0 x x 0 x x 0 f(x) g(x) By the limit laws, 0 0 x x 0 x x 0 f(x 0 ) f(x) g(x) Also, 0 x x 0 x x 0 b = f (x 0 ) lim x x 0 (x x 0 ) x x 0 (f(x) g(x)) = f(x 0 ) g(x 0 ), so a = f(x) f(x 0 ) b(x x 0 ) f(x) f(x 0 ) b = f (x 0 ) b, so x x 0 x x 0 x x 0 x x 0 So we see that the linearization is the only possible g(x), as claime We can use the linearization of a function to n numerical approximations to function values The iea is to n a function f, a nice value of x 0, an a small x such that f(x 0 ) is easy to compute an f(x 0 + x) is the value we want Then f(x 0 + x) f(x 0 ) + f (x 0 ) x gives the approximate value esire Example: Fin a linear approximation to f(x) = ln(x) near x = 1, an use it to approximate the value of ln(09) For f(x) = ln(x) we have f (x) = 1/x Thus, the linearization near x = 1 is L(x) = ln(1) + (x 1) 1 = x 1 Then ln(09) L(09) = 01 With a calculator, we compute ln(09) 010536, so the linear approximation is quite goo Example: Approximate the values of 11 an 14 To approximate 11 the most natural thing to try x 0 f(x 0 + x) = 11 is the value we want to compute = 1 an x = 01, an f(x) = x: then For f(x) = x = x 1/2 we have f (x) = 1 2 x 1/2 We get the approximation 11 = f(1 + 01) 1 + 1 2 1 1/2 (01) = 1 + 1 01 = 105 2 22
With a computer, we obtain 11 = 104881 So we can see that the linear approximation is quite goo If instea we take x = 04, we get the approximation 14 = f(1 + 04) 1 + 1 2 1 1/2 (04) = 12 With a computer, we obtain 14 = 118322 The approximation is still goo, but not as goo as the previous one Remark: There is no reason only to consier approximation by linear functions, asie from the fact that linear functions are the easiest to analyze We coul just as well look for quaratic approximations a 0 + a 1 x + a 2 x 2, or cubic approximations a 0 + a 1 x + a 2 x 2 + a 3 x 3, or general polynomial approximations to f(x) This is precisely the iea behin Taylor polynomials, which (among other things) give even better approximations of a function than the tangent line oes However, eveloping these ieas fully requires integral calculus, so we will not pursue the iscussion further at the moment It is also possible to use ierent classes of functions to approximate functions, such as trigonometric functions Approximating a function by one of the form a 0 +a 1 sin(x)+b 1 cos(x)+a 2 sin(2x)+b 2 cos(2x)+ leas to the iea of a Fourier series, which possess an equally eep an useful theory (an which, unfortunately, we also will not evelop further at the moment) 272 Dierentials Denition: When y = f(x) is given as a function of x, the ierential y is ene as y = f (x) x, where x is an inepenent variable As can be seen by rearranging the enition, we have ene ierentials so that the erivative, which we have enote as f (x) = y, is now actually a quotient of variables x The intuitive iea of ierentials is that as x tens to zero the approximate equality y f (x 0 ) x becomes more an more accurate In the limit we turn the symbols into symbols Dierentials are a piece of what is sometimes (pejoratively) calle abstract nonsense: some theoretical notation which is useful an yiels correct answers, but whose rigorous use we o not have the tools to justify at this stage There are several ierent technical ways to make sense of ierentials, but all of them involve signicantly more avance mathematics than calculus Our primary interest in ierentials is that they greatly simplify some computations an arguments in integral calculus, particularly in calculations involve in changing variables Dierentials also simplify an make clearer some of the properties of erivatives: for example, the Chain Rule's ierential form z x = z y y can now be interprete as a cancellation of the y terms x Example: If y = x 2, then we have the ierential y = f (x) x = 2x x Example: If y = x 3 an z = y 2, n z in terms of x By the enition we have z = 2y y an y = 3x 2 x Plugging the secon into the rst gives z = 2(x 3 )(3x 2 x) = 6x 5 x Alternatively, we coul have written z = y 2 = (x 3 ) 2 = x 6, an then irectly compute the ierential z = 6x 5 x Note that we obtaine the same result either way, on account of the Chain Rule Well, you're at the en of my hanout Hope it was helpful Copyright notice: This material is copyright Evan Dummit, 2012-2016 You may not reprouce or istribute this material without my express permission 23