2.4 Exponential Functions and Derivatives (Sct of text)

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1 2.4 Exponential Functions an Derivatives (Sct of text) 2.4. Exponential Functions Definition Let a>0 be a real number ifferent tan. Anexponential function as te form f(x) =a x. Teorem (Exponent Properties). Let a>0 be a real number ifferent tan. If x an y are any real numbers, ten. a 0 = 2. a = a 3. a x a y = a x+y a x 4. a = y ax y an a = a x x 5. (a x ) y = a xy. Teorem (Important Limits). Let a>0 be a real number ifferent tan.. If a>, ten(seefigure) (a) lim a x = x! (b) lim x! ax =0 (c) lim a x = lim =0. x! x! a x 2. If 0 <a<, ten(seefigure2) (a) lim a x =0 x! (b) lim x! ax = (c) lim a x = lim =. x! x! a x Figure : Grap of te function f(x) =2 x Figure 2: Grap of te function f(x) = 2 x Teorem Te exponential function f(x) =a x is eiter an increasing or a ecreasing function wit omain R an range (0, ). Moreover, a x > 0 for all x in R an te function f(x) =a x is one-to-one. 68

2 2.4.2 Derivatives of Exponential Functions Recall te efinition of te erivative of a function. Definition Let f be a function. Te function f 0 efine by te formula f 0 (x) =lim f(x + ) f(x) is calle te erivative of f wit respect to x. Using te above efinition, we sall compute te erivative of te exponential function f(x) =a x, were a>0 is a real number ifferent tan. First, we state an important teorem. Teorem Let a>0 be a real number ifferent tan. Ten te limit exists. a lim Let f(x) =a x.applyingteefinitionofteerivativeoff wen x =0,weobtain Terefore, f 0 (0) = lim f(0 + ) f(0) a =lim. f 0 a (0) = lim. (?) Teorem Let f(x) =a x, were a>0 is a real number ifferent tan. Ten Proof. Since f(x) =a x,weobtain Terefore, f 0 (x) =f 0 (0)a x. f 0 f(x+) f(x) (x) = lim f 0 (x) =f 0 (0)a x. by efinition of erivative. = lim a x+ a x since f(x) =a x. a = lim x a a x = lim ax (a ) property of exponents. istribution property. = a x a lim limit property. = a x f 0 (0) by (?) above. 69

3 2.4.3 Te Natural Exponential Function a Tere is only one real number a suc tat lim =. We sall give tis number te special name e. Te real number e is irrational an is approximately Itisavery important constant, just as is a important constant. Definition (Te number e). Te number e is te unique real number satisfying e lim =. Te function f(x) =e x is calle te natural exponential function. Teorem Let f(x) =e x. Ten lim f 0 (x) =e x. Proof. Let f(x) = e x. By Teorem we see tat f 0 (x) = f 0 (0)e x. Since f 0 (0) = e =,weavetatf 0 (x) =e x. Remark. Using a ifferent erivative notation, te above teorem states tat x [ex ]=e x. Teorem (Cain Rule Revisite). Let g(x) be a ifferentiable function, ten e g(x) = e g(x) g 0 (x). x Problem. Differentiate te following functions:. f(x) =e 3x+x3 2. f(x) =x 2 e x2. 3. f(x) = x2 e x2. 4. f(x) =(e 2x + x 2 ) 2. Teorem 2.4. (Exponent Properties). If a an b are any real numbers, ten. e 0 = 2. e = e 3. e a e b = e a+b 4. e a e b = e a b an e a 5. (e a ) b = e ab. = e a 70

4 Teorem Te exponential function f(x) =e x is an increasing function wit omain R an range (0, ). Sof(x) =e x is a one-to-one function. Also, e x > 0 for all x in R. 20 y y x 0 x Grap of te natural exponential function f(x) =e x Teorem (Limit Properties). Te following ol:. lim e x = x! 2. lim x! ex =0 3. lim e x = lim =0. x! x! e x Logaritmic Functions Recall tat te exponential function f(x) =a x is eiter an increasing or a ecreasing function wit omain R an range (0, ). Tus, a x > 0 for x in R an te function f(x) =a x is one-to-one. Consequently, te function f(x) =a x as an inverse function f (x), wic we call te logaritmic function wit base a an is enote by log a (x). Sinceterangeof f(x) =a x is (0, ), it follows tat te omain of log a (x) is (0, ). Definition Let a>0 be a real number ifferent tan. Te logaritmic function log a (x) is efine by (see Definition on page 65) y =log a (x) if an only if a y = x. Problem 2. Write eac in exponential form:. 5=log 2 (32). 2. 2=log(00).[Note:log is unerstoo to mean log 0.] 3. 0=log e (). 4. c =log e (a). Remark Applying Definition on page 65, we see tat. a log a (x) = x for all x in (0, ); 2. log a (a x )=x for all x in R. 7

5 Te next two teorems ientify te important properties of logaritmic functions. Teorem If x an y are positive numbers an r is any real number, ten. log a () = 0 2. log a (xy) =log a (x)+log a (y) 3. log a ( x y )=log a(x) log a (y) 4. log a ( y )= log a(y) 5. log a (x r )=r log a (x). Teorem (Limit Properties). Let a>. Te following ol:. lim x! log a (x) = 2. lim x!0 + log a(x) = Te Natural Logaritmic Function Recall tat te natural exponential function f(x) =e x is an increasing function wit omain R an range (0, ). Tus, e x > 0 for x in R an te function f(x) =e x is one-to-one. Consequently, te function f(x) =e x as an inverse function f (x), wic we call te natural logaritmic function wit base e an is enote by ln(x). Since te range of f(x) =e x is (0, ), it follows tat te omain of ln(x) is (0, ). Definition Te natural logaritmic function ln(x) is efine by (see Definition on page 65) y =ln(x) if an only if e y = x. Remark Applying Definition 2.3.9, we see tat. ln(e x )=x for all real numbers x. 2. e ln(x) = x for all positive real numbers x. 3. ln(a) =b if an only if e b = a. 4. ln(e) =. Teorem (Limit Properties). Te following ol:. lim x! ln(x) = 2. lim x!0 + ln(x) = y 3 2 y ln x 0 x Grap of te natural logaritmic function f(x) =ln(x) 72

6 2.4.6 Derivative of te Natural Logaritmic Function Teorem Let f(x) =ln(x). Ten Proof. Since x [ln(x)] = x. e ln(x) = x (2.) for all positive real numbers x, we sall ifferentiate bot sies of te equation (2.). Using te cain rule on te left sie, we obtain e ln(x) [ln(x)] =. x Tus, solving for x [ln(x)] we obtain (recall tat eln(x) = x) x [ln(x)] = e = ln(x) x. Teorem (Cain Rule). Let g(x) > 0 be a ifferentiable function, ten x [ln(g(x))] = g0 (x) g(x). Problem 3. Differentiate te following functions.. f(x) =ln( p x). (Tis makes sense, provie x>0.) 2. f(x) =ln(x 2 x 2). (Tis makes sense, provie x 2 x 2 > 0.) 3. f(x) =(ln(3x)) f(x) = x ln(x). 5. f(x) =e x ln(x). 6. f(x) =ln(x 2 + e x ). Problem 4. Fin te equation of te tangent line to te function f(x) =6x te point (, 6). ln(x 2 ) at Corollary Let f(x) =ln( x ). Ten f 0 (x) = x. Proof. Tere are two cases to consier: Case : x>0. In tis case we ave tat f(x) =ln( x ) =ln(x) an ence f 0 (x) = x by Teorem Case 2: x<0. In tis case we ave tat x>0 an f(x) =ln( x ) =ln( x) an ence f 0 (x) = x by Teorem Terefore, f 0 (x) = x = x. So in eiter case we ave tat f 0 (x) = x. 73

7 2.4.7 Derivatives of General Exponential an Logaritmic Functions Teorem Let a>0 ifferent tan. Ten a x = e x ln(a) (2.2) an (a x ) 0 =ln(a) a x. (2.3) Proof. Let a>0. Ten, recalling Remark 2.4.9() an Teorem 2.4.6(4), we see tat a x = e ln(ax) = e x ln(a) an tus, using equation (2.2). (a x ) 0 =(e x ln(a) ) 0 =ln(a)e x ln(a) =ln(a) a x Problem 5. Let f(x) =x + x. Evaluate f 0 (x). Solution. Note tat is a constant. Observe tat in te expression x,tebaseisavariable an te exponent is a constant. Tus, we use te power rule to obtain (x ) 0 = x.onte oter an, x is an exponential function, because te base is a constant an te exponent is a variable. So ( x ) 0 =ln( ) x by Teorem Terefore, Problem 6. Let f(x) =x x. Evaluate f 0 (x). f 0 (x) =(x + x ) 0 = x +ln( ) x. Solution. Observe tat in x x,tebaseanteexponentarefunctionoftevariablex. Since x x = e ln(xx) = e x ln(x), we see tat f 0 (x) =(x x ) 0 =(e x ln(x) ) 0 = e x ln(x) (x ln(x)) 0 = e x ln(x) x x +ln(x) Teorem Let a>0 be a real number ifferent tan. Ten log a (x) = ln(x) ln(a) = e x ln(x) ( + ln(x)). an x [log a(x)] = ln(a) x. Proof. Let y =log a (x). Tus, a y = x. Hence,ln (a y ) = ln(x) an so, y ln(a) = ln(x) 74

8 by a property of logaritms. Solving for y we obtain y = ln(x) ln(a). Because y =log a(x) we now conclue tat log a (x) = ln(x) ln(a). Tus, x [log a(x)] = ln(a) x. Problem 7. Let f(x) = log 0 (x 2 +). Evaluate f 0 (x). Solution. Using Teorem an te cain rule, we obtain f 0 (x) = ln(0) (x 2 +) (x2 +) 0 = 2x ln(0) (x 2 +). If f(x) =log a (g(x)), ten(inasimilarmanner)onecansowtatf 0 (x) = Logaritmic Differentiation g0 (x) ln(a) g(x). Te labor of ifferentiating a function f(x) involving quotients, proucts, or powers can often be substantially reuce by first applying te function ln an using its properties, togeter wit te Cain Rule. Proceure. Let f(x) =g(x), were g(x) is a formula for f. To fin f 0 (x), bylogaritmic ifferentiation, o as follows:. Take ln of bot sies of te equation f(x) =g(x), obtainingln(f(x)) = ln(g(x)). 2. Expan ln(g(x)) an write te equation ln(f(x)) = your expansion. 3. Differentiate (wit respect to x) bot sies of te equation obtaine in step Solve te equation obtaine in step 3 for f 0 (x), replacingf(x) wit its formula g(x). 2)(x 2 +).Finf 0 (x) using logaritmic ifferenti- Problem 8. Let f(x) =(x 2 +3x)(x ation. Solution. Since f(x) =(x 2 +3x)(x logaritmic function, we see tat 2)(x 2 +),usingtealgebraicpropertiesoftenatural ln(f(x)) = ln(x 2 +3x)+ln(x 2) + ln(x 2 +). Tus, by ifferentiating bot sies of te above equation an using te cain rule, we obtain f 0 (x) f(x) = 2x +3 x 2 +3x + x 2 + 2x x 2 +. (2.4) Solving equation (2.4) for f 0 (x) an recalling tat f(x) =(x 2 +3x)(x 2x +3 f 0 (x) =f(x) x 2 +3x + x 2 + 2x x 2 + =(x 2 +3x)(x 2)(x 2 +) 2x +3 x 2 +3x + x 2 + 2x x 2 + 2)(x 2 +),weobtain. 75