Rules of Differentiation

Transcription

1 LECTURE 2 Rules of Differentiation At te en of Capter 2, we finally arrive at te following efinition of te erivative of a function f f x + f x x := x 0 oing so only after an extene iscussion as wat te rigt an sie of tis formula means, an ow to compute it. But now, as promise, we are going to evelop tools tat will allow us to compute erivatives witout actually taking its. Rater we sall evelop rules by wic we can compute erivatives from our knowlege of a few basic erivatives. Te first step will be to evelop a repetoire of basic erivatives. However, before we even start, let me write own an alternative it efinition of te erivative. Suppose we want te value of te erivative of f at a particular point x = a. Ten f a + f a a = x 0 Now instea of tinking of going to 0, let us tink instea tink of x := a + going to a. So oing we can write, =, an Tus, f a + f a f x f a a = = x 0 f x f a a := x is an alternative, but equivalent, efinition of te erivative of f at x = a. Okay, now we re reay to evelop te basic rules of ifferentiation. We ll start wit rules to compute te erivatives of simple powers of x. 0.. Te Power Rule. Let s o te easiest case first. Suppose f x = for all x. Ten we ave Next consier f x = x n. In orer te compute tis it we note tat f x + f x x = = = 0 = 0 x x + n x n x = x 0 A n B n = A B A n + A n B + A n 2 B AB n 2 + B n 58

2 2. RULES OF DIFFERENTIATION 59 Tis can be seen by expaning te rigt an sie an noticing tat all but te first an last terms en up cancelling RHS : = A B A n + A n B + A n 2 B AB n 2 + B n We tus ave = A A n + A n B + A n 2 B AB n 2 + B n B A n + A n B + A n 2 B AB n 2 + B n = A n + A n B + A n 2 B A 2 B n 2 + AB n A n B A n 2 B 2 + AB n B n x + n x n x xn = 0 x + n + x + n x + x + n 2 x x + x n 2 + x n = = = x + n + x + n x + x + n 2 x x + x n 2 + x n n=0 0 Applying formula, we get x + x x + n + x + n x + x + n 2 x x + x n 2 + x n = 0 Simplifying te first factor in te numerator, we get x + n + x + n x + x + n 2 x x + x n 2 + x n = 0 Cancelling te common factor of from te numerator an te enominator, we get 2 = x + n + x + n x + x + n 2 x x + x n 2 + x n 0 Now we re taking te it of a polynomial an tis we can o by simplying evaluating eac term at = 0 because polynomials are always continuous. In fact, we can o tis term by term. So we en up wit x xn = x + 0 n + x + 0 n x + + x + 0 x n 2 + x n 3 = x n + x n + + x n + x n Te only ting tat remains is figure out exactly ow many x n terms appear on te rigt an sie. Looking back at te original expansion we see tat we a a total of n suc terms, for note tat in 2 we ave factors of x tat range from = x 0, x, x 2,..., x n, an so we re basically counting from 0 to n, wic will ave te same number of steps as counting from to n. Conclusion: we ave a total of n terms x n on te rigt an sie of 3. Tus, we can finally conclue: In fact, we ave Teorem 2.. Suppose α is any real number, ten x xn = n x n x xα = αx α Example 2.2. Suppose f x = x π/2, wat is f x? π x xπ/2 = x π/2 2

3 2. RULES OF DIFFERENTIATION Te Constant Multiple Rule. Suppose f is a ifferentiable function an c is a constant. Ten cf x + cf x f x + f x cf x = = c = cf x x 0 0 were in te secon equality we ave use te Limit law cf x = c F x wenever c is a constant. Tus, we ave prove. Teorem 2.3. Suppose f is a ifferentiable function an c is a constant, ten cf x = c x x x Example 2.4. Compute te erivative of f x = 3x 5. Well, 3x 5 = 3 x 5 = 3 5x 5 = 5x 4 x x were we ave employe bot te Constant Multiple Rule an te Power Rule Te Sum an Difference Rules. Recall tat F x ± G x = F x ± G x was anoter of our basic Limit Laws. From tis law we see tat, if f an g are ifferentiable functions, One similarly sows tat Tus, f x + + g x + f x + g x f x + g x = x 0 f x + f x g x + g x = = g x + x x x g f x g x = x x x x x Teorem 2.5. Suppose f an g are ifferentiable functions, ten g f x ± g x = x ± x x x x Example 2.6. Compute te erivative of f x = 3x 2 + 2x Well, we ave x = 3x 2 + x x 2x x = 3 2x x 0 = 6x + 2 Remark 2.7. We often call tis process ifferentiating term-by-term.

4 2. RULES OF DIFFERENTIATION Derivatives of Exponential Functions. Recall tat an exponential function was efine as a function of te form a being some constant. f x = a x We can actually compute te erivatives of suc functions. For Notice tat wen x = 0 we ave a x+ a x x = = a x a n x x 0 0 x 0 = a n a n a0 = 0 0 Tus, in general, if f x = a x is an exponential function x = ax f 0 = f x f x Definition 2.8. Suppose e is te unique number suc tat Ten te exponential function e = 0 f x = e x is calle te natural exponential function. We often use te notation for tis exponential function. exp x := e x Te reason wy tis f x = e x is more natural from te point of view of Calculus tan oter exponential functions like f x = 0 x is tat x ex = e x It is, in fact, te unique function suc tat f 0 = an f x = f x. In summary, Teorem 2.9. x ex = e x Example 2.0. Compute te secon erivative of f x = 3x 2 + 4e x. Well, te first erivative of f is x = 3x 2 + 4e x x = x = 3 3x x ex by te sum rule x x2 + 4 x ex by te constant multiple rule = 3 2x e x by te rules for ifferentiating x n an e x = 6x + 4e x

5 2. RULES OF DIFFERENTIATION 62 Te secon erivative is te erivative of te first erivative, so 2 f x 2 : = = x x x 6x + 4ex = 6 x x + 4 x ex = 6 x + 4e x = 6x 0 + 4e x = 6 + 4e x 0.5. Te Prouct Rule. Recall tat te rule F x G x = F x G x mae it easy to compute te its of proucts of functions. Unfortunately, tis rule can not be applie irectly to te computation of erivatives of proucts of functions. For on te one an 3 Wile on te oter an g x x x x f x + g x + f x g x f x g x := x 0 : = 0 f x + f x 0 g x + g x = 0 f x + g x + f x + g x f x g x + + f x g x wic oes not agree wit te rigt an sie of 3. To get te rigt rule, let me start wit te alternative formula for te erivative f a We ten ave f x f a a = x f x g x f a g a fg a = x f x g x + 0 f a g a = f x g x + f a g x f a g x f a g a = g x f x f a + f a g x g a = f x f a = g x + f a Now we can apply te prouct rules to bot its = g x f x f a + f a Now g x = g a since if g x is ifferentiable it is also continuous. Also f a = f a because f a oesn t epen on x. Tus, we get f x f a fg a = g a x + f a g x g a g x g a g x g a = g a f a f a g a

6 2. RULES OF DIFFERENTIATION 63 Now regaring te evaluation point a as a free parameter x, we get Teorem 2. Te Prouct Rule. Suppose f an g are ifferentiable functions. Ten x fg x = f x g x + f x g x Remark 2.2. Tis rule is also calle te Leibniz rule since it was first publise by Gottfrie Leibniz in Te Simple Quotient Rule. From our it laws we also know tat P x = P x so long as Q x 0 Q x Q x But just as wit te Prouct Rule above, we ave to be a little careful an getting te correct rule for ifferentiating quotients of functions. We ll start wit a particularly simple quotient of functions. Suppose f x = Qx. Ten f x f a a : = = x Q x Q a Fining a common enominator for te secon factor, we ave a = x Q a Q x Q x Q a Ten, collecting factors ifferently an using te Prouct Rule for Limits we ave 5 x a = Q a Q x Q x Q a Employing te Quotient Rule we ave Q x Q a = Q x Q a = Q a 2 so long as Q a 0 an, from te efinition of te erivative, we ave Q a Q x = Q x Q a = Q x We can tus conclue from 5 tat x a = [Q a] 2 Q x Regaring te evaluation point a as a variable x, we tus arrive at te following rule: Teorem 2.3 Te Simple Quotient Rule. Suppose Q x is a ifferentiable function. Ten x Q x = Q x [Q x] 2 so long as Q x 0. At points x were Q x = 0, te erivative of Qx nee not be efine.

7 2. RULES OF DIFFERENTIATION Te General Quotient Rule. Let us now consier a function of te form f x = P x /Q x. Wit our spanky new Prouct an Simple Quotient Rules in an, we can reaily erive a formula for te erivative of P x /Q x. For all we ave to o is tink of P x /Q x as P x /Q x. Tus, P x = P x x Q x x Q x P = x Q x + P x x Q x = P x Q x + P x Q x Q x 2 We tus ave = P x Q x P x Q x Q x 2 Teorem 2.4. Suppose P x an Q x are ifferentiable functions. Ten P x = P x Q x P x Q x x Q Q x 2 at all points x suc tat Q x 0. Example 2.5. Compute te erivative of f x = x2 3x + x We ave f x of te form P x Qq wit P x = x 2 3x + Q x = x To apply te quotient rule we nee to compute P x = x 2 3x + = 2x 2 3 x + 0 = 2x 3 x Q x = x 3 + = 3x = 3x 2 x We can now plug into te quotient rule x = x P x Q x = P x Q x P x Q x Q x 2 = 2x 3 x 3 + x 2 3x + 3x 2 x 3 + 2