Derivatives. if such a limit exists. In this case when such a limit exists, we say that the function f is differentiable.

Transcription

1 Derivatives 3. Derivatives Definition 3. Let f be a function an a < b be numbers. Te average rate of cange of f from a to b is f(b) f(a). b a Remark 3. Te average rate of cange of a function f from a to b is te slope of te cor joining te points (a, f(a)) an (b, f(b)). We are intereste in stuing te limit as a an b are close togeter so tat te previous cor becomes a tangent. Remark 3.3 Consier a car moving in a lane. Let f(x) be te istance of te car from te initial position after x secons. Ten, te average rate of cange of f from a to b is te average spee of te car witin te uration starting from te a t secon an ening at te b t secon. Definition 3.4 Let f be a function an x be a number. Te erivative of te function f is a function, enote by f, so tat f f(x + ) f(x) (x) = lim, 0 if suc a limit exists. In tis case wen suc a limit exists, we say tat te function f is ifferentiable. Remark 3.5 Anoter common notations for te erivative of f at x is f (x) = x f(x). Sometimes, people present a function as y = f(x) an tey write its erivative as x. Tese alternative notations are prove to be convenient wen we encounter cain rule later. Definition 3.6 Let f be a function an a be a number. Te tangent line (or just tangent) to te grap of f at a is te straigt line wit slope f (a) wic passes troug te point (a, f(a)).

2 Example 3.7 Evaluate te erivative of te function f(x) = x for all numbers x 0 For x 0 an 0 < < x, So, for all x 0. f (x) = lim f(x+) f(x) = [ x+ x ] = x (x+) x(x+) = x(x+). 0 x(x + ) = x Example 3.8 Evaluate te erivative of te function For x > 0 an 0 < < x, So, for all x > 0. f(x) = x for all numbers x > 0. f(x+) f(x) x+ x = = ( x+ x)( x++ x) ( x++ x) = (x+) x ( x++ x) = ( x++. x) f (x) = lim 0 ( x + + x) = x Example 3.9 Evaluate te erivative of f at 0 if For 0, f(x) = x for all numbers x f(0 + ) f(0) =. But lim 0 oes not exist (wy?). So, te function f oes not ave a erivative at 0. Tat is, f is not ifferentiable at 0. Teorem 3.0 A ifferentiable function is continuous. ifficult

3 3. Stanar Formula for Derivatives Teorem 3. If a is a number an f(x) = a for all x. Ten, f (x) = 0. For 0 an all x, So, f(x + ) f(x) = a a = 0 f (x) = lim 0 0 = 0. Teorem 3. If n is a positive integer an f(x) = x n for all x. Ten, f (x) = nx n. For 0 an all x, So, f(x + ) f(x) = (x + )n x n = (x + ) n + (x + ) n x + (x + ) n 3 x (x + )x n + x n. f (x) = lim 0 (x+) n +(x+) n x+(x+) n 3 x +...+(x+)x n +x n = nx n. Teorem 3.3 If f an g are ifferentiable functions, (f + g) = f + g. For 0 an all x, So, f (x) (f + g)(x + ) (f + g)(x) f(x + ) + g(x + ) (f(x) + g(x)) = f(x + ) f(x) g(x + ) g(x) = + f(x + ) f(x) = lim + 0 f(x + ) f(x) = lim 0 = f (x) + g (x). g(x + ) g(x) + lim 0 g(x + ) g(x) 3

4 Teorem 3.4 If f an g are ifferentiable functions, (f g) = f g. omitte En of lecture 9 Teorem 3.5 If f is a ifferentiable function an a is a number, (af) = a(f ). For 0 an all x, So, (af)(x + ) (af)(x) a(f(x + )) a(f(x)) = f(x + ) f(x) = a (af) (x) f(x + ) f(x) = lim a 0 f(x + ) f(x) = a lim 0 = a(f (x)). Example 3.6 Let f(x) = + x + 3x + 4x 3 for all x. Evaluate te erivative of f. Te erivative of is 0 Te erivative of x is times te erivative of x wic is Te erivative of 3x is 3 times te erivative of x wic is 6x Te erivative of 4x 3 is 4 times te erivative of x 3 wic is x So, te erivative of f is f (x) = + 6x + x. Teorem 3.7 (Prouct Rule) If f an g are ifferentiable functions, (fg) = f g + fg. For 0 an all x, (fg)(x + ) (fg)(x) f(x + )g(x + ) (f(x)g(x)) = f(x + )g(x + ) f(x + )g(x) + f(x + )g(x) f(x)g(x) = g(x + ) g(x) f(x + ) f(x) = f(x + ) + g(x) 4

5 Now f(x + ) f(x) lim 0 an lim 0 g(x + ) g(x) exist an are f (x), g (x) respectively, an lim 0 f(x+) = f(x) as f is continuous (follows from te ifferentiability of f), (fg) (x) g(x + ) g(x) = lim f(x + ) 0 g(x + ) g(x) = lim f(x + ) lim 0 0 = f(x)g (x) + f (x)g(x). f(x + ) f(x) + g(x) f(x + ) f(x) + g(x) lim 0 Teorem 3.8 (Quotient Rule) If f an g are functions, ( f g ) = gf fg g. Since f = f g. Taking erivatives on bot sies of te equation by prouct rule g yiels Rearrange te equation an we get f = ( f g ) g + f g g. ( f g ) = gf fg g. Example 3.9 If f(x) = for all x. Evaluate te erivative of f. x7 Te erivative of f is, by te quotient rule, f (x) = (0)x7 ()7x 6 (x 7 ) = 7 x 8. Remark 3.0 In a similar way, we are able to sow tat if n is a negative integer, te erivative of f(x) = x n is f (x) = nx n. Example 3. Fin te line tangent to te grap of f(x) = at te point x3 (, ). Since f (x) = ( 3)x 4 = 6 x 4, we ave f () = 6. Te line tangent to te grap of f at (, ) is te straigt line wit slope f () = 6 wic passing troug (, ). Tat is, te line wose efining equation is y = 6(x ) or y = 6x

6 Example 3. Let a > 0, an C is te grap of f(x) = x /4a. Fin te line tangent to C at (at, at ). Hence sow tat tis line makes te same angle wit te y-axis an te line joining (0, a) an (at, at ). Since f (x) = x/a for all x. Te line tangent to C at (at, at ) as slope t an so its efining equation is y at = t(x at), or y tx + at = 0. Now te line joining (0, a) an (at, at ) as slope (at a)/at = t/ /t. Terefore, tangent of te angle subtene by tis line te tangent line in te first paragrap is t (t/ /t) + t(t/ /t) = t wic is te same as tangent of te angle subtene by te line tangent to C at (at, at ) an te y-axis. Teorem 3.3 If f(x) = sin x for all x. Ten, f (x) = cos x. For 0 an all x, f(x + ) f(x) sin(x + ) sin x = = sin(x + + ) sin(x + ) So, = sin(x + ) cos + cos(x + ) sin sin(x + ) cos + cos(x + ) sin = cos(x + )sin. f (x) = lim cos(x + 0 )sin = lim cos(x + 0 ) lim 0 = cos x sin Teorem 3.4 If f(x) = cos x for all x. Ten, f (x) = sin x. similar Teorem 3.5 If f(x) = tan x for all x. Ten, f (x) = sec x. 6

7 Since tan x = sin x. We take erivative by quotient rule an get cos x f cos x cos x + sin x sin x (x) = cos = sec x. x Teorem 3.6 If f(x) = cot x for all x. Ten, f (x) = csc x. Since cot x = cos x. We take erivative by quotient rule an get sin x f sin x sin x cos x cos x (x) = sin = csc x. x Teorem 3.7 If f(x) = sec x for all x. Ten, f (x) = tan x sec x. Since sec x =. We take erivative by quotient rule an get cos x f 0 cos x + sin x (x) = cos = tan x sec x. x Teorem 3.8 If f(x) = csc x for all x. Ten, f (x) = cot x csc x. Since csc x =. We take erivative by quotient rule an get sin x f 0 sin x cos x (x) = sin = cot x csc x. x Definition 3.9 Te limit a lim 0 is increasing in a. It is 0 wen a = an te limit is big provie tat a is sufficiently large. Define e to be te number suc tat e lim =. 0 Teorem 3.30 If f(x) = e x for all x. Ten, f (x) = e x. f (x) = lim 0 f(x + ) f(x) = lim 0 e x+ e x e x e e x = lim 0 = e x e lim 0 = e x. 7

8 En of lecture Cain Rule Teorem 3.3 (Cain Rule) Let f an g be functions. Ten, proof (sloppy): For all numbers x an 0, Tus, (f g) (x) = f (g(x))g (x) f g(x + ) f g(x) f(g(x + )) f(g(x)) = f(g(x) + (g(x + ) g(x))) f(g(x)) = g(x + ) g(x) g(x + ) g(x) (f g) (x) f(g(x) + (g(x + ) g(x))) f(g(x)) g(x + ) g(x) = lim 0 g(x + ) g(x) f(g(x) + (g(x + ) g(x))) f(g(x)) g(x + ) g(x) = lim lim 0 g(x + ) g(x) 0 f(g(x) + k) f(g(x)) g(x + ) g(x) = lim lim k 0 k 0 = f (g(x))g (x). Remark 3.3 Te cain rule can be presente alternatively as follows: Let u = g(x) an y = f(u) (a way to present te composite function f g.). Te cain rule becomes x = u u x. Example 3.33 Let y = (x + ) 4. Evaluate x. Let u = x +, ten y = u 4 so tat On te oter an, u = x + so tat u = 4u3. u x =. 8

9 Combining tese results by cain rule, we ave x = (4u3 )() = 8(x + ) 3. Alternatively, we may let g(x) = x +, f(x) = x 4. Ten, f g(x) = (x + ) 4 an (f g) (x) = g (x)f (g(x)) = ()(4(g(x) 3 )) = 8(x + ) 3. Example 3.34 Let y = sin(x + ). Evaluate x. Let u = x +, ten y = sin u so tat On te oter an, u = x + so tat By cain rule, = cos u. u u x = x. x = u u x = x cos u = x cos(x + ). Example 3.35 Let y = xe x. Evaluate x. y is naturally te prouct of x an e x. By prouct rule, x xex = x x ex + e x x x. We know tat x =. So, te issue is to compute x v = e u. By cain rule, Terefore, x ex = v u u x = eu = e x. x xex = xe x + e x = e x (x + ). x ex. Let u = x an Example 3.36 Let p an q be integers, f(x) = x p/q. Evaluate f (x). Let y = x p/q. Ten, y q = x p. We ave x yq = x xp. 9

10 Te rigt an sie is known to be px p. Moreover, x yq = ( x )( yq q ) = qy x. Terefore, or q qy x = pxp x = p x p q y q = p q x p q. Remark 3.37 From te previous example, we see tat te erivative of f(x) = x k is f (x) = kx k for all rational numbers k. In fact, suc formula is true for all real numbers k. Example 3.38 Fin te erivative of y = cos e ex. Let v = e x, u = e v so tat y = cos u. Ten, x = u v u v x = cos u e v 3.4 Implicit Differentiation e x x u v = ( sin u)(e v )(e x ) = e x+ex sin e ex. Definition 3.39 An implicitly efine function is a function efine by specifying a relation between x an f(x) for every number x. Example 3.40 A function y is implicitly efine by Evaluate te erivative of y. x 3 + xy(x) + (y(x)) 3 =. Take erivatives on bot sies of te given equation, we get 3x + y + x x + x y3 = 0. We may moify te last term by cain rule, 3x + y + x x + 3 x = 0. 0

11 Tat is Rearranging yiels En of lecture 3x + y + x + 3y x x = 0. x = + y 3x x + 3y. Example 3.4 Let a > b > 0 an C be te curve collecting all points (x, y) suc tat x a + y b = (*). Fin te line tangent to C at ( a b, ). Verify tat a ( a a ) + b ( b a ) =. Te given point ( a b, ) is on te curve C. Consier C (or part of it) as te grap of a function y. Tis function is efine implicitly by (*). We ifferentiate bot sies of (*) an get x a + y b x = 0. Rearranging yiels x = b x a y. Tus, te slope of tangent of C at ( a, b ) is x = b x a y (x,y)=( a, b ) = b a. Consequently, te line tangent to C at ( a, b ) as efining equation y b = b a (x a ) or bx + ay = ab. Example 3.4 Let f be a function suc tat Evaluate f (). f(x)e f(x) = x for all x.

12 Differentiate bot sies of te given relation, we obtain Tus, f (x)e f(x) + f(x)f (x)e f(x) = for all x. f (x) = e f(x) + f(x) for all x. Observe tat e =. Terefore f() = an ence f () = e f() + f() =. Teorem 3.43 Let f be a one-to-one function wit inverse g. Ten, g (x) = /f (g(x)) for all x unless f (g(x)) = 0. Since g is te inverse of f, f(g(x)) = x for all x. Differentiate bot sies of te equation an apply cain rule, we get f (g(x))g (x) =. Tat is, wenever f (g(x)) 0. g (x) = f (g(x)) Remark 3.44 If te function is presente as y = f(x), te inverse function is presente as x = g(y). Te previous teorem is saying tat te erivative of g, tat is, x is relate to te erivative of f by x =. x Were te functions are to be evaluate at te appropriate points. Teorem 3.45 If f(x) = ln x for all x > 0. Ten, f (x) = x for all x > 0. Let g(x) = e x an f(x) = ln x, ten f (x) = g (f(x)) = e = f(x) e ln x = x. Example 3.46 Let y = e x x x sin x. Evaluate x.

13 Since y = e x x x sin x y = e x x x sin x ln y = ln(e x x x sin x) ln y = ln e x + ln x x + ln sin x ln y = x + x ln x + ln sin x, we take erivative on bot sies of te equation an get x ln y = + ln x + x x + cos x sin x ln y = + ln x + cot x x = + ln x + cot x y x x = ( + ln x + cot x)y x = ( + ln x + cot x) e x x x sin x. Example 3.47 Evaluate te erivative of y = tan x. Since y = tan x, tan y = x an π/ y π/. Hence, x tan y = x x tan y = x sec y x = x = sec y x = + tan y x = + x. Example 3.48 Let y = sin x. Evaluate x. 3

14 Since y = sin x, sin y = x an π/ y π/. Hence, x sin y = x x sin y = x cos y x = x = cos y x = sin y x = x. En of lecture since π/ y π/ an tus cos y Higer Derivatives Definition 3.49 Let f be a function, te erivative of f is calle te secon erivative of f. Its symbol is f. Te erivative of f is calle te tir erivative of f. Its symbol is f. Similarly, te n t erivative of f is te erivative of te (n ) t erivative of f an te symbol for te n t erivative of f is f (n). Remark 3.50 If a function is presente as y = f(x), its n t erivative is also written as n y x n. Example 3.5 Evaluate te secon erivative of f(x) = e x. By cain rule, Ten, by prouct rule an cain rule, f (x) = xe x. f (x) = e x + 4x e x = e x ( + 4x ). Example 3.5 Evaluate te first seven erivatives of f(x) = 4x 4 +3x 3 +x +x. f(x) = 4x 4 + 3x 3 + x + x f (x) = x 3 + 9x + 4x + f (x) = 36x + 8x + 4 f (x) = 7x + 8 f (4) (x) = 7 f (5) (x) = 0 f (6) (x) = 0 f (7) (x) = 0. 4

15 Remark 3.53 If f is a polynomial of egree, te n t erivative of f is 0 wenever n >. 3.6 Rates of Canges Definition 3.54 Suppose tat tere is a particle moving in a line. Define f(t) =istance of te particle from a certain fixe point on te line at time t. Ten, f (t) is calle te velocity of te particle at time t an f (t) is calle te acceleration of te particle at time t. Example 3.55 A particle is moving in a line suc tat it is locate x(t) = e 3t sin 4t meters away from its initial position after t secons. Evaluate its initial velocity an initial acceleration. Since we ave x (t) = 6e 3t sin 4t + 8e 3t cos 4t x (t) = 4e 3t sin 4t 48e 3t cos 4t x (0) = 8 x (0) = 48. Tus, te initial velocity of te particle is 8ms, its initial acceleration is 48ms. Definition 3.56 Wen tere is is a quantity canging over time an f(t) is tat quantity at time t. f (t) is calle te rate of cange of te given quantity at time t. Example 3.57 In a cemical reaction camber, te concentration of a certain reactant A after t secons is [A](t) = M. Here k an c are two fixe k(t + c) positive numbers. Sow tat te rate of cange of te concentration of A is proportional to te square of te concentration of A. Te rate of cange of te concentration of A (rate of reaction) is Tus, te assertion is prove. t [A] = t k(t + c) = k(t + c) = k[a]. 5

16 Example 3.58 Water is pumpe into an inverte conical vessel wit semivertical angle π/6 at a rate of R m 3 s. Evaluate te rate at wic water level in te vessel is raising at te moment wen te water level is at H meters. Let V (t) be te volume of water in te vessel after t secons. (t) is te water level in te vessel after t secons. Ten, V (t) = 3 (t)π((t) tan π 6 ) = π 9 (t)3 for all t. Taking erivatives on bot sies of te equation yiels V (t) = π 3 (t) (t). Suppose tat after t 0 secons, te water level in te vessel is H meters. Tat is, (t 0 ) = H. It is given tat V (t) = R for all t. In particular, V (t 0 ) = R. We evaluate te previous equation at t 0 an it yiels R = π 3 H (t 0 ) or (t 0 ) = 3R πh. 3R Te water level is raising at a rate of πh ms at te moment wen te water level is H meters. Example 3.59 One is keeping an eye on a car wic is moving in a straigt line. At te moment wen te car is closest to te guy, te car is D meters away from te guys an it is moving at a spee of V meters per secon. Evaluate te rate at wic te guy turns is ea at tat moment. Let O be te point at wic te car is closest to te guy. x(t) is te istance between te car an O after t secons. θ(t) is te angle tat te line joining te guy an O makes wit te line joining te guy an te car. Ten, x(t) = D tan θ(t) for all t. Taking erivatives on bot sies of te equation yiels x (t) = Dθ (t) sec θ(t). In oter wors, θ (t) = D x (t) cos θ(t). Suppose tat after t 0 secons, te car is locate at O. Tat is, x(t 0 ) = 0 an θ(0) = 0. Evaluating te last line at t 0 yiels θ (t 0 ) = D x (t 0 ) cos θ(t 0 ) = V D. 6

17 Consequently, te guy as is ea turning at V D raians per secon wen te car is closest to im. Example 3.60 Air is pumpe into a sperical balloon at a rate of R m 3 s. Evaluate te rate at wic te surface area of tis balloon is canging wen its surface area is S 0 m. Let V (t), S(t), r(t) be te volume, surface area an raius of tis balloon respectively after t secons. Ten, Eliminating r(t) yiels Differentiate bot sies an we obtain V (t) = 4π 3 (r(t))3 S(t) = 4π(r(t)) for all t. V (t) = 6 π S(t)3/ for all t. V (t) = 4 S(t)S (t) for all t. π Now let te surface area of te balloon be S 0 after t 0 secons. Ten or so tat V (t 0 ) = 4 S(t0 )S (t 0 ), π R = 4 S0 π S (t 0 ) π S (t 0 ) = 4R. S 0 Consequently, te rate at wic surface area of te balloon is canging is π 4R m s. S 0 En of lecture 3 7