THE DERIVATIVE Definition of the Derivative @ Using the Definition to Calculate Derivatives @ Derivatives of the Basic Functions © Derivatives of Sums, Products, Quotients, and Composite Functions @ Derivatives Involving Repeated Applications of the Product, Quo and Chain Rules % The Differentiation Formulas: Using the Chain Rule @ Higher Order Derivatives 4-1. Definition of the Derivative ‘The derivative of a function y = f(x) isa function f°(x) defined as a limit: THE fix + Ax) — fQ) DERWATWVE $0) = is « where Ax (read “delta x") represents a change in x. Your textbook may use the equivalent definition ; flx+ hy — f0) = fim JE = J ea eae The domain of f is the set of all numbers for which the limit exists; at those values of x, f is said to be differentiable. ‘The process of finding a derivative is known as diferentation ‘The derivative of y = f(x) is denoted in several ways: dy ca d oD z e j fo OF DS EF Efe) yor The presence of x in D,f, dy/dx, and djjdx is to remind you that x is the independent variable (and you speak of the derivative with respect to x). If ¥ = f(0),then you would write dy/dr, df jdt, and so on and speak of the derivative with respect to t. Yousnay want to find the valu of the derivative fora particular value of x. The value of f'(x) at x = ais denoted (as you might guess) by f(a, dy F@) o 2 42. Using the Definition to Calculate Derivatives You can use the definition of the derivative to find the derivative of simple functions, EXAMPLE 4-1: Differentiate f(x) = x 6h62 Caleulus Solution: First find the difference quotient f(x + Ax) ~ F009 Ax and simpliy it algebraically: fle + Ax Ax x? + 2x Ax + (Ax)? 7 Ax __ Axx + Ax) Ax ‘Then apply the definition of the derivative Se + Ax) — fe) LOe nae = fn 22x £49 “yo Ax 1 lim (2x + Ax) = 2x aro . EXAMPLE 42: Differentiate f(x) = 7° ‘Solution: Proceed as before: ‘foe + Ax) =f) _ Te + Ax — Tx? ‘Ax ax [x3 + 3x2(Ax) + 3x(Ax)? + (Ax?) — 7x? . ax __ TB? + 3x(Ax) + (8x) JAX - Ax and so im £0. + Ax) = fox) £0) = Ii ano ax om ‘T[Bx* + 3x(Ax) + (Ax)*] Ax Axo Ax Jim 7[3x? + 3x(Ax) + (Ax)? ao = 21x? EXAMPLE 43: Differentiate f(x) = 1/x. ‘Solution: Construct the difference quotient: toot fly + Ax) — fx) _ x Ex x Bx 7 Be x- (e+ Ax) xe + Ax) Ax ope © (Ax)x(x + Ax)‘The Derivative 63 ‘Then take the limit: jim Gs aro Qaae + x) -1 -1 Fey © gto xe + A) You should begin to see that using Definition 4-1 to find derivatives has definite disadvantages: The algebra can be tedious and the calculation of the limits may be nearly impossible. To avoid repetition of these calculations, you apply the definition of the derivative to a generic function of a particular type in order to develop a general formula. EXAMPLE 4. Solution: The Differentiate f(x) = x", where n isa positive integer. rence quotient is Le + AX) = fo) _ (x + Ax ~ Ax ‘Ax Use the binomial formula fe paste me ty Dae ey ay {o simplify this quotient, The difference quotient becomes Sle + Ax) ~ fx) _ x" + mx Ax) + dnl — Dx 2x? 4 + (Any! Bx Ax Ce! + dln = 1)x72(Ax) 4 + (Ans Je wear aa as and so Fa) = tim UL + mn ~ De “ao fo $ (Ax) Ax r = lim [nxt" + dnln — Dxt3(Ax) + (A acto = mt You have a formula! ot 4 Rea for na postive integer EXAMPLE 4-5: rentiate y = x” with respect to x. ‘Solution: Use the formula from Example 4-4 with n = wa dx dx («’) xT = 76 4-3. Derivatives of the Basic Functions If you apply the definition of the derivative to the basic functions, you will find the following formulas. Its very important that you familiarize yourself with this list, d d ropes 1") = att Je for any constant ¢ 3. FG") = me! for any real number m of) forany d 4. Flbin x) = cos x constant ¢64 Calculus 5. (os) = sin x 6“ (ams) = sectx a 1. 4 (eee) = seo x tan x . Gees) = d 8, F(osox) = —ese xeot x az = —cse?: jy ot) = — ose Ae bera > 0 10. @'In@ for any num- a ne Re a 1 2. = Los] = pg for a>Oa¢l a 13, Fins EXAMPLE 46: Differentiate Solution: Simply use formula 2.and then formula 3: a d 3) ROE Note: Replace radical signs with fractional exponents when using the formulas. EXAMPLE 4-7: Differentiate y = /x. Solution: a du 1 yaa RW) = Rea pe EXAMPLE 48: Differentiate y = 4 /x*. Solution: a ax Recall that 1/(x" formulas, EXAMPLE 49: Differentiate y = 2/x°. Solution: ‘7x? with respect to x (0) = 73x81 = 2? Bg) wg ot xo" You may use negative exponents when using the (69) = 2-3 = et 1s ween) a 16, 5 (are tan x) = a 17, Fare see x) = 18, Awe) = 20, Jann x} = cosh x a , 21. “(cosh x) = sinh ae n ¢ (tanh x) = sech?x a 23, 7 (seeh x) = —sech x tanh x 24, 4 (esch x) = eich x cath x & # eoth x) = ~csch?x a Ta 1 ae 20 3 3‘The Derivative 63 EXAMPLE 4-10: Differentiate y = 3/,/3°. Solution: 45) fa) = 3.4 tu-8y = af 3) gtrstr-t ROVE) = 3-F ) = af —F]atsy -1 x EXAMPLE 4-11: If f(x) = Jx/ 9%; find /'(). Solution: Express f(x) asa power of x xt Se) = Spy = 82-009 = 6 as fe) = rey = 20 4-4, Derivatives of Sums, Products, Quotients, and Composite Functions Most ofthe functions you'll have to differentiate won't be as simple as those listed in Section 4-3. Instead, you'll encounter sums, products, quotients, and i composite functions, A. Derivative of a sum The derivative of the sum (or difference) of two or more functions is the ‘sum (or difference) of the derivatives of the functions. PATE d nee Re F/G) + obo = s'0) + 9) 43) Asan immediate result ofthis formula and the first three formulas of Section 4 4-3, you can now find the derivative of a polynomial EXAMPLE 4-12: Differentiate y = 2x? — 4x? + Tx 4 6, Solution: The derivative of a sum is the sum of the derivatives: fae at =f ary — 4 ays 4 cd Be Or — 4 +8 +O = FAH) — FO) + 28) + 1) The derivative of a constant times a function is the constant times the derivative of the function, so. 4 in igd, & Ott EO = 42x) + 711) +0 x? Bx +7 EXAMPLE 4-13: Differentiate y = 4x? + 12x5 — ldx? + Tx — 5. Solution: Dx 4x5 + 1265¢4 — 429 47-0 a = 28x° + 60x — 28x +7 EXAMPLE 4-14: Diflerentnte f(x) = Sx? + x + (I/x) — GV).66 Caleutus ‘Solution: First express the function in a form that allows you to apply the formulas: Sl) = 5x x pct = a Now differentiate: Sx) 41 + (ttt = Qt = 158 $b fe) B. The product rule The derivative of the product of two functions f(x) and g(x) ‘f(3) times the derivative of g(x) plus g(x) times the deri ‘Symbolically, (in words) e of f() pmonucr 4 Ffeietall = fal) + abot) 4 EXAMPLE 4.15: Differentiate y = (x3 + 3x — I)? — Solution: a a al? + 3x et — x — 1) noted nd ers ae a = (8 $ 3x =O + 3x DFO = (0 + 3x — 1Qx = 1) + (x? = x — Gx? + 3) EXAMPLE 416: Differentiate y = x? sin x. Solution: 4 (sin xy = 2 4 gy let sing) = tina) + six 009) i ? cos x + sin x (2x) = x? cos x + 2x sin.x EXAMPLE 4-17: Differentiate f(x) = (cos x)/x", Solution: = x7 ¥(~sin x) + ¢08 x(—3x™ C. The quotient rule ‘The derivative of the quotient of two functions fs) rey wheres) #0 is more easily expressed symbolically than it is with words: uonient 4 (fls)) _ aleif'x) — fists) rs) UE a \g09 toto? However, it will help you to remember the formula if you verbalize it as follows: The derivative of the quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all divided by the square of the denominator. a 2 4 i 1 ‘D. The chain rule Youll use the chain rule to find the derivative of a comps functions, flax) ‘Note: Your text may use the notation (g° f)(x) to describe composite functions. Simply remember that (a° le) = lf) A. The first difficulty you'll encounter is recognizing a composite expression ‘when you see one. EXAMPLE 4.21: Consider the function h(x) = (3x — 2). This function is the omposition of two functions g(x) = 3x — 2 and f(x) = x4,s0 Gs) = f(g) hink about the steps you'd perform to evaluate (3x — 2)* at x = 5. First youd find3 x 5 — 2 = 13,and then you'd raise 13 to the fourth power, Thus, the inner fUnction is 3x — 2 and the outer function isthe fourth power function, x* __ EXAMPLE 4.22: Consider the function f(x) = sin(x? + 1). This function is @ Sompostion of sin x and x? + 1. To find /(2) you'd fist find 2? + 1 and then ‘Ply the sine function to 5. Thus, the inner function is x? + 1 and the outer “unction is the sine function. ‘The Derivative 6768 Calculus 2. The symbolic statement of the chain rule for f(g(x)) is CHAIN d " “ ure Fla) = Paed)ere 4) Ifyou think of f(g(x)) asa composition of outer function f(x) and inner function g(x), then the derivative of the composition is the product of the derivative of the outer function, evaluated at the inner funtion, and the derivative of the inner function. EXAMPLE 4-23: Differentiate h(x) = (x — 2)*. Solution: First recognize that h(x) = f(g(x)), where the outer function is f(x) = x4, and the inner function is g(x) = 3x — 2. The derivative of the outer function is fey = 4x3 So the derivative of the outer function, evaluated at the inner function, is F(alx)) = 4{g(x))* = 48x ~ 2)° ‘The derivative of the inner function is ge) By the chain rule, the derivative of h(x) is 4 H() = F*(g)a'(x) = 4x — 293) = 12x — 29° EXAMPLE 4-24: Differentiate h(x) = sin(x? + 1) Solutio atx) The outer function is flx) = sinx; the inner function is ? + 1. The derivative of the outer function is J"(x) = cos x,80 The derivative of the inner function is g'(x) = 2x, so AY) = '(gOd}g'(2) = oos(x? + 1)2x 2x cos(x? + 1) EXAMPLE 4-25: Differentiate y = (x7 + 5)° Solution: The outer function s x’; its derivative is 3x2, Thus, the derivative of the outer function, evaluated at the inner function, is 3(x? + 5)?. The derivative of the inner function is simply 2x, so oe 3x? + 5)°2x) = Gabe? + 5)? EXAMPLE 4-26: Differentiate f(x) = e*7'. Solution: ‘The inner function is 5x — 1; the outer function is the exponential function, e%. The derivative of the outer function is e*, which yields e*”" when evaluated at the inner function. The derivative of the inner function is 5, so £09) eg) = set EXAMPLE 4-27: Differentiate f(x) = In(2 + sin x). Solution: The outer function is In x; the inner function is 2 + sin x. Fe) eee of outer function (derivative of inner function) evaluated at inner function 1 TF \) )The Derivative 69 EXAMPLE 4-28: For f(x) = (x! + x)°,find f'(). FiG0) = xt? + ve dein +x) = 304? + ae + :) SA) = 34"? + a(t + ‘) pti)=ns peas = 3044) (5 EXAMPLE 4-29; Differentiate f(x) = sin e Solution: Note that this is a composition of more than two functions, but don't panic: It simple requires more than one application of the chain rule, Let the notation guide you. At each stage, simply identify the outer function. The outer function is sin x; the inner function is e**?*"5, Thus, me 7 arr) To find (d/dx\(e***?*"*), use the chain rule a second time, with outer funetion e and inner function x? + 2x + 5: ceri Lies) amnermnfetenete sas] SS EXAMPLE 4-30: Differentiate y = cos*(x* + 1), Solution: The outer function is x*, and the inner function is cos(x* + 1), 30 § (eosi(y? + I) = Scost(x? + (eae +H) & ‘Now the outer function is cos x and the inner function is x* + 1, s0 dy Fe = Sos? + IE—singx? +e eet) = —Seost(x? + Nsin(x? + N[3x*] = —15x?eos(x? + Isin(x? + 1) 4-5. Derivatives Involving Repeated Applications of the Product, Quotient, and Chain Rules Lets consider a few problems that, at first glance, look more difficult than those We've considered so far. Although you must use more than one rule for each expression, you won't need any new rules or techniques. EXAMPLE 4:31: Differentiate F(x) = xPe* — [6 + 2x — x)] + sin(x? + 1). Solution: First recognize f(x) as a sum of derivatives: f= + + (sin? + 1)) ‘Then use the product rule on the first term, the quotient rule on the second term,70 Caleulus and the chain rule on the last term: SO) = x (z «) + ee x) wy Aeta— +9£e-x) 2 EXAMPLE 4-32: Differentiate f(x) = sin (5 = fe In ) Solution: 1n this example, you will find compositions, quotients, and products of functions, It isnt necessary that you envision all of the rules and steps al once to differentiate the function. Rather, you can differentiate the function one step at a ime, You must merely recognize the next step, letting the notation guide you. The function fis @ composition of sinx (the outer function) with [(? + 2x)/(x — 1)] In x (the inner function). Apply the chain rule as your first step toward finding /"(x): a © sin x = cos x a sede \ d (xt + 2x Inx)— Inx a&\x-7 Now to differentiate [(x? + 2x)/(x — 1)] In x, you use the product rule, as thisis* a product of two functions: 1 : So = oo(S +2 ) f $F A in +n »EE3)] ‘To differentiate (x? + 2x)/(x ~ 1) you must employ the quotient rule, as this isa quotient of two functions: F'(3) = 608 & S a In ) [5 = i. (:) £8) = cos| &- pders 2x) = (x? + 2x). fit) | + (nx) ay ox(S $e), ) fe +e ay y coe oe? +2) EXAMPLE 4-33: Differentiate f(x) = x°/n(e* +1. Solution: The first feature of f(x) that you must recognize is that f(x) is a quotient. Accordingly the first step toward finding /"(x) is an application of the quotient rule: Ine* + Ae - edie +1) Sy= Finke?‘The Derivative 71 In(e* + 1)3x? Se = wea) 3x7 In(e* + 1) — Serle" + 1) Wwe + i) _ PXAMPLE 4-34: Differentiate /(x) = Sin veos x 2D. Solution: You can replace the square root sign with an exponent. Now recognize as a composite function with outer function x" and inner function “sin xc0s x + 2 S(s) = (sin x cos x + 2)" f= 5 sin x cos + 384 Gin osx +2 Now use the product rule: sO) = Sin eos x 2M [sn leas 3) + cos x4 a qin) + x0 = Sein x 008 + 2)-"fsin x(—sin x) + 608 x(G08 x) + 0] 4 (sin x cos x + 2)” "? (cos?x — sin?x) EXAMPLE 4-35; Differentiate fx) = /x? + 3x +1 Je) — 1, Solution: Rewriting with fractional exponents, : Sls) = (2? + 3x 4 1!G9 — 19 yousee that f(x) isa product. Your first step then is an application of the product rule: a a Co) = (2 43x 4 eas — 8 4 03 — yee ye I 8) = (0? + 3x + 1) Be O— P + GP ~ DI Ge + 3x +1) Each of the terms to be differentiated now requires the chian rule: Sx) = (7 + e+ oly (1 ce - 4 +0 [jos +3x+1) koe +3xt »| (x7 + 3 + 1) — Gx) 3 + oe — 1G? + 3x + MAK + 3) +6. The Differentiation Formulas: Using the Chain Rule ' Rename the inner function, Hy = f(g(s)), then the chain rule says d) z = (derivative of f, evaluated at g) (derivative of g)72 Calculus For notational purposes, rename the inner function g(x) = u. Then y = f(u)and the chain rule becomes dy py du ET IOER EXAMPLE 4-36: Differentiate f(x) = sin yx. Solution: Let u = J = x*?, the inner function, Then fix) = sin w £0) = cos ut = cos x Ger) x B. The chain rule applied to the basic functions Reexamine the list of derivatives in Section 4-3. You will find these formulas following, written ina different notation. Each function is a composition of the given outer function with an inner function w dy at tt a tu 1 tut) = mE formany WF (eosin) = Fm Sk real itumber A u i. a 13, 5 (arweos w) = 2. ZGsinw) = cos ue : a du 14, © (are tan y) = 3. loos) = sin & d 18, Flare see u) a >, ft 4, Gy (tan w) = sec’ “: d d _ du 16. — (are ese u) = Gq bora) = secu tan & a —1 dw d du = Ao 6. 5 (ose) = exc ucot u 7 17. 5 (are cot) = aa a du 7. £ (cot a) = ~esctu 18, Gin a) = cosh u | a du | don wnat! 2 ae & Ewa Ina forany 19. F [eosh w) sinh uF number a > 0 P ™ 20, 4 tanh v) = sechu a du & & 9. See dx dx. d du ‘21, —— (sech u) = —sech u tanh u— to dogey = (2) a de rn 78) = \iina de a is a>0a4l 2 a oso a) = ~esch woth a ues a 2, du ; 2, otha) = ~ csc j These formulas are not an added complication. They are simply the same formulas you learned in Section 4-3, ‘with the chain rule “built in.” Remember that ts ropresents any differentiable function of x. If for example, w= x, then dujdx = 1 and these formulas become those you learned in that earlier section | :EXAMPLE 4.37: Differentiate y = sin(1/x). Solution: Formula 2 tells us that (gin Aon - So, with w= 1/x “O EXAMPLE 4-38: Find J tanh xe) ‘Solution: Apply formula 20, where the inside function isu = xe": a 2, de (anh) = seoha ad a (ian xe") = S soh?ee) A (ee) = sech*(xe*}(xe* + e*) Note that you needed the product rule to find (d/dx) xe" 4-7. Higher Order Derivatives ‘The second derivative of afunction y = f(x)is the derivative of the first derivative of f The second derivative of y = f(x) is denoted in a variety of ways: ay ef a fo) DY DF TE Gr Gall) yor EXAMPLE 4-39: Find the second derivative of f(x) = 2x — 4x? + 7x + 6. Solution: f'(e) = 10x4 — 8x 47 f= “(pon) = 4 ox - 8x +7) = 40x? -8 dx. Ix. ‘The nth derivative of a function y = f(x) (where mis a positive integer) is the derivative of the (n — 1)st derivative of f. The nth derivative of f is most often denoted (x) to avoid the use of many primes. EXAMPLE 4-40: Find the sixth derivative of f(x) = 2x5 ~ 4x? + Ix + 6. Solution: Ox = 8x +7 fe) 7-8 SMR) = 120x2 F(x) = 240% FONx) = 240 (9) = 0 LO) EXAMPLE 4-41: Find the higher derivatives of f(x) = sin x. Solution: F') = 008 x f"G) = —sin x FX) f(x) = sin x cos x So the pattern continues. ‘The Derivative 734 Calculus 1. The derivative ofa function is another function, which is defined as the limit of a difference quotient. 2. In practice, you differentiate the basic functions by using the formulas in Section 4-3. Itis important that you know these formulas. 3. You may have to express the function in another form to use the formulas. If a function involves radicals, express them as fractional exponents. 4. Use the product, quotient, or chain rules to differentiate functions involving | products, quotients or compositions. | 5, You must cortectly identify composite functions to use the chain rule 6. You can differentiate a complicated function one step ata time, The notation will guide you | 7. To calculate the mth derivative of a function, differentiate times. PROBLEM 41 Use the definition of the derivative to find f(x), wherg f(s} = 3x ~ 4 Solution: First find the difference quotient and simplify: ‘ fle + Ax) ~ fle) _ (Boe + Ax) = 4] ~ Bx - 4) ax Ax ese rs Beta To find the derivative, take the limit as Ax approaches zero: Fea = lim (2) =3 [See Section 42] PROBLEM 4-2. Use the definition of the derivative to find f(x), where f(x) = 3 ~ 2x ~ 2x’ Solution: Find the difference quotient and simplify: floc + Ax) — fla) _ [3 — 28 + Ax) — 20 + Ax) ~ [3 — 2 = 27] Ax ~ Ax 4xdx — 2(Ax - 3 + 2x + 2x7 Ax _ 3 = 2x = 2Ax - 2x? 2x — dex — 20x)? ax Apply the limit to find f(x) —2Ax — 4xAx — 2(Ax) “@) = |i Eiaaee eee) a) sn ( ae ) fim A252 = 48 = 289) yay {Se Section 42) ano ax PROBLEM 43. Use the definition of the derivative to find "(x), where fx) = 3/(2x ~ 1), Solution: Find the difference quotient: 3 3 flx + Ax) = fle) _ 200 + Ax) AeThe Derivative 75 Find a common denominator and simplify Sle + Ax) — f(x) _ 32x - 1) ~ 32x + 24x - 1) ‘Bx © AxQx = Qe + 2x — I __ bax © AxQe = Dx + 2Ax — 1) ‘Apply the limit: fl Sie + Ax) — fe) _ —6Ax PO) = fim ae ase Oe FAD oa "oF (See Scion #2) PROBLEM 4-4 Find the derivative of f(x) = 1/,/'x, using the definition of the derivative, Solution: Find the difference quotient and rationalize the numerator: fle + x) ~ fo) _ hfe FB) ~ (thy) Ax ax Ax /x,/x + Ax a Ws = Vet ane + Ve Fae) 4 (axe fx + Axil + /e + Ax) a“ x= (e+ Ay) Ax Jala + Axx + Je + Ax), eee Ax Sele + Axil + Je + AN) Apply the limit: 4 S69) = lim ee aeadx Sale + Axil + yx + Ox) 1 = lim RT ge EE 1 et * VeBile+ Pa” x25) [See Section +2) PROBLEM 4.5 Use the definition of the derivative to find (x), where Ss) = 2x? + x? - Sx 43, Solution: Sle + Ax) — fo) _ (Ge + Ax)? + (& + Ax? ~ S(x + Ay) + 3] - Qe? +37 - Se +3) ax = ax = Ale + Ax)? = 9] + [le + Ax) ~ x7] — SE + Ax) - x) oe eee ees _ 203x7Ax + 3x(Ax)? + (Ax) + [2xAx + (Ax}*] - Sx - Ax Axf6x? + 6xAx + 2(Ax)? + 2x + Ax — 5] —ee eeere- 16 Calculus von yom LE # AN) = fe) foe ‘ax tim SXU6H? + OxAx + 2x + 2x + A ao ax lim [6x? + 6xAx + 2(Ax)? + 2x + Ax ~ 5] aro = 6x? + 2x - 5 [See Section 42] PROBLEM 4-6 Find the derivative of flx) = 3x¢ + 2° — 2x? ~ dx #5. ‘Solution: The derivative of a sum is the sum of the derivatives: pe) = Lawt + fee) Ea 5+ do and the derivative of a constant times functions the constant times the derivative of the fanction: 7 4 ya 4 oy S10) = 304) + EO 2) = 3(4x') + 3x? = 20x) — A(t) +0 21d tbe dx 4 (Se Section 44] PROBLEM 4-7 Find dyjde where y= —2x7 + 12x — Ux? + 3x j 2 Solution: You proceed as in the last problem: 5 aay yt aid (+ FOAL + 30+ RO = =2(7x8) + 12x) = 1G?) + AY +0 = —14x® + 48x? - 33x? + 3 [See Section 44] PROBLEM 4-8 Differentiate y = (2x°/7) + (6x7/9) + 2x/5) ~ 1 § Solution: Because the denominators are constants, you don’t need the quotient rule: &y (2) 4 ya) 40+ (4 mea? # 2) £0r+()i d+ (sla) 2a 2 =(j) +(S) Ports Se Section 4 tty [See Section 44] PROBLEM 4-9. Differentiate f(x) = 88°" + (2/) ~ GB/s’) Solution: First write each of the terms as a power of x Se) = 8x59 + 2x? © py a9 (2x80) + edt 2 3 49 as — 2x7? + 6x7? [See Section 4-4} PROBLEM 4-10 Differentiate y = eye. Solution: This function can be written as a power of x: x8 SDS [See Seation 64]The Derivative 77 PROBLEM 4-11 Differentiate f(x) = (x? + x °Y(x? ~ 3). Solution: Because this is a product of two functions, you should use the product rule: sesorsryt wate 9 bee a) a = (x8 + x 9)Qx) + (0? — 3)Bx? — 3x4) [see Seton 4-4) PROBLEM 4-12 Differentiate f(x) = x°e" Solution: Use the product rule: FG Je + e fey xe* + e%(3x7) [See Section 4-4} PROBLEM 4-13 Differentiate f(s) = ¢ arc tan Solution: Use the product rule d a og are tan + are tan (9) fas t = 75 + aretant ce Section 4 tert [See Section 44] PROBLEM 4-14 Differentiate y = (2x + 5)/(x — 3) Solution: Because this is a quotient of two functions, you use the quotient rule: a d dy OOD GEA I~ O47 -9 dx (3 _ (= 3)2) ~ Ox + SK) & 3? and finally simplify dy = oF AAs an alternative, you can write y as @ product and use the produet rule: ay od eg lx + He - 3) d = ad HOt ITO“ YH I'L OS) To differentiate (x ~ 3)"', you use the chain rule; Deas 9| -Hs - yee - >| +6 -3°Q) a = Qx + M(x — 377] + 2 - 3)" _rOxt5 2 5+ 20-3) “=a Te 3 > (= 3)? = “@-F You can use either of these methods to differentiate a quotient—it's up to you. (See Section 44 PROBLEM 4.15 Differentiate y = (I + tan x)/sin x.78 Caleulus Solution: Use the quotient rue d a { ay at + tan x) = (1 + tan) 7 G6in) f dx (Ginx? _ sin x(0 + sec?) ~ (I_+ tan x)o0s x jin x sec?x — ¢08 x sin’ [See Section 44 PROBLEM 416 Differentiate f(x) = (x* — 3x + 2)"7 Solution: Because this function is the composition of two functions, x* — 3x +2, the inner function, withthe sevententh power function (x) the outer function, you can use the chain rule: £ (x) isthe derivative of x", evaluated at x — 3x + 2, times the derivative of x* ~ 3x + 2 P(X) = 1T08 = 3x + 2 Ee —3x4+2) = 115 ~ 3x + 2)16(5x4 — 3) {See Section 44] PROBLEM 417 Differentiate y = sin’ x, Solution: You'll recognize y asa composite function if you rewrite it as y = (sin x)°. Now yousee that yis@ composition of sin x, the inner function, with x°, the outer furition. Use the chain rule: = ain xp? 4 a bin 2) = 3sin? x 608 x [see Section 4.4] | a 1 1 PROBLEM 4-18 Differentiate f(x) =e" "**"! i | i Solution: Thisis a composition of the exponential function, the outer function, withx? + 3x + 1, the inner function, so Fy = rears por tixt =e ey + 3) [Soe Section #4] PROBLEM 4-19 Differentiate y = log,(x + 2x + 3). ; Solution: Use the chain rule: dx x 4 e+ 3\in3 1 ax +2 -sernlm)®-@rarms snes PROBLEM 4-20 Differentiate f(x) = 3/(x? + 1). Solution: The easiest way to differentiate this function is to rewrite it as f(x) = 30x? + 1) use the chain rule: a $0) = MO? + Ete? +) 6x = 30? +170) = Sp You could also differentiate f using the quotient rule: @ +ofo—38 foe +) Ot F _9= 308) bx "EP + TP f(x) = {See Section 4-4]The Derivative 79 PROBLEM 4-21 Differentiate f(x) = In(t + sin®x), Solution: Use the chain rule: 1 [4a oe 10=am [get | fae OT¥s (sin x 208 x) [See Section 4-4] PROBLEM 4-22 Differentiate y = (x? + x — 3)!(2x3 — Sx? + 9x + Vf Solution: Use the product rule: a a 24x — 32 toge — 52 4 (xd — 5x2 4 tay yh (TOE WPT Oe ~ Sa? + 9 + E+ Ox? — x9 4 Oe + EO? + x — 3)! = (x7 + x ~ 3)!24(2x9 — Sx? Ox + ytor — 5x? + 9x +1) 3 5x2 8 12(x? 14 a | 4+ Ox? — Sx? 4+ 9x + D2 $x — 3)! TO? + x = 3) | = (2? + x = 3MIx — Sx7 4 Hx + NOx? — 10x + 9) 4 + (2x? = Sx? + 9x + I}*12G0? + x — 32x + 1) . You may want to simplify this expression: ‘ | YL ae Hae — $e 5 Fe 7 MOP te DEAN — 5x7 + 9x +O) i x [2? + x — 3)(6x? — 10x + 9) + 32x? — Sx? + 9x + NE +] = AG? + x — Ax? — Sx? + 9x + 1) (184 — 28x4 + 20%? + 72x — 24) = B(x? + x — I Qx3 — Sx? + 9x + 1)(9x4 — 14x3 + 10x? + 36x — 12) [See Sotion 453 ; PROBLEM 4-23 Differentiate f(x) = In(log,» x). Solution: Use the chain rule: d ibkex lz (logio »] aA [See Section 4 ~ jogo x \x/ in 10 aes PROBLEM 4-24 Differentiate f(x) = a4, Solution: Use the product rule: (a) = xt A? — gy 4g 2 — aye Ly $18) = PF = 4) + 6 = aA ty _ ode a] ee? atten a = 522 — 47109) + tet — 4p? Simpti [See Scion 45] PROBLEM 4-25 Differentiate f(x) = e™"*®,i 80 Calculus i i ‘Solution: Use the chain tule: ' i F | voy = 0 ging? J) gin? + D) Be ees? + L(t + er & it = ef *Mfcos(x? + 1)](2x) {See Section 45.) PROBLEM 4-26 Differentiate f(x) = \/(? + Die — 5), ‘Solution: You can differentiate f(x) = [(x? + 1)/(x — 5)]" using the chain and quotient rules: fh afer a t x5) a a a yen ro( ~ 9268 OF HEU °) 2\x—5 e-5P _ esty Ge — 92x) — 02 FY] Aft #1) 1x ~ 10 “x= 5 ee x5 (x — 5) You can also differentiate f by rewriting it in the form, \ 241 Fo) = (2 + Ge — 3 and using the product formula, {See Section 45} PROBLEM 4:27 Differentiate f(x) = sectr/x? = 1). Solution: Use the chain sule: f(0) = seole = Dann (P= net fF = sets? = Ttanle fa Dr[ S00 . pede - o| = soot /X? — Ntanin/ 1) 8 [See Seaton 45] PROBLEM 4-28 Differentiate f(x) = cosh 2x sinh x. Solution: Use the product rule: (0) = cosh 2x“ (sinh ») + sinh x 4 (cosh 23) a & = cosh 2x cosh x + 2sinh x sinh 2x [See Seton 45] PROBLEM 4-29 Find the third derivative of f(x) = 4x9? — 3xt 4 x? — 1 \ ‘Solution: Differentiate three times: S's) = 6x! — 12x + 2x F(x) = 3x"? — 36x? +2 fie)= fre =x (see Sein 47) PROBLEM 4-30 Find the second derivative of y =‘The Derivative 1 clution: Dilferentiate twice: dy a Dd secx =2 = 2sec?x Ge 7 280% F (6002) = 2 see x see tan x = 2 sec?x tan x 2 ayo a 2, sects F (tan x) + 2tan.x 7 (e0?s) at = 2sectx sect + (2 tan x)(2 see x}(sec x tan x) = Dsectx + 4 see?x tan? {See Section 47) PROBLEM 4-31 Find the sixth derivative of f(x) = In(t + 2). ‘Solution: Dillerentiate six times: f= = =(L+a! (= (+ xr? s(x) = YL + xy? SQ) = —6( + y-* Fx) = 24 + > SMR) = — 1201 + x)-# [See Section 4.7) Supplementary Exercises In Problems 4-32 through 4-41 use the definition of the derivative to find the derivative of f(x) 432 f(x) = 5x44 437 f(x) = 42-4 433 fix)= 22 4x1 438 f(x) = x + (I/x) 4340 f(x) = 3x? - 12x +1 4:39 f(x) = /x — 2x 435 f(x) = 1/(e + 1) 440° f(x) = Bx — N+ xT 436 f(x) = Je 7 441 f(x) = 20? +) In Problems 4-42 through 4-127 find the derivative of the given function, 4-42 f(x) = 3x3 + Sx? — Tx + 450 f(x) = 3sinx + 4tanx 443 f(x) = x ~(I/x) 451 y= xsinx 4644 f(x) = Sx? ~ 13x + 6 452 y= xe + 3x 44 445 yore dbs tore 483 y = arcsinx + arccos x 445 y =e 454 fla) = (2 + 1" ~ 1) 447 y= Selle 455 y= (x + Sox - JX) 448 fla) = 8 JPY 456 y= x/lnx 449 f(x) = x? 4 2 457 f(x) = (are sin x)/x82 Caleulus 458 f(x) = (In 2)/In x 459 f() = (8 - 3x +7) x (5 — 2Y% + 3x) 460 ya(Ptx—N/G-% 4461 y = L/(l + cosx) 462 y = sin x/(sin x + 008 x) 4463 fix) = 0° +! 464 foy=tye-4 465 fx) = Wy) + Je #2 466 f(x) = Sin 4-65 y = In(x? + 2) 4-68 y = tan'x 469 y=ennn 4m y= Vite an y= Jeeu—5 4:72 y = cosh(sinh x) an y=e? 414 fix) = 27 4:75 f(x) = sin(2") 4.16 f(x) = sin®2 417 fix) = Sie? 2) 4:18 y = [Sin ¥ 008 x 479 y= 2" 480 y = coi(I/x) 481 y=(e 4x 3? 482. y = arcsec dx sin xe" 485 f(x) = x7 Insin x FL + x} 487 f(x) = 2x8? — Sxt 4 Tx? 1 488) = Gee 489 y = (2x? + 3x + AY[O - ST 490 y= J@= DTD 491 y= 10x - = 1) 492 f(x) = cos TF 493 f(x) = Ox? - 11)? — 3) 4.94 f(x) = Ine + JT+) 493. fix) = (8 — Me +) 4.96 y = [InQx ~ 1) 497 y= Inin(x + 4) i 498 y = logs (/x { x) 499 y = 10" logo x 4100 y= PTT 4 4-101 y = [isin x)/x]° 4-102 y = (Inxjin(x + 1) rm | 4-103, y = Infx in(x + I] 4-104 y = tan*(In x) 4108 y = cxce* +e 4-106 y = In'(2 + sinx) 4407 y = Inf(x' sin x? + 1) 4-108 y =e 4109 y =e" 4-110 f(x) = sin(sin x) + sin?x 4-111 f(x) = In + e) 4112 y = tanh(3 ~ 5x) 4413 y = Vlogs * 4d fo) = fe tye eT 4415 f(x) = (5/6) JT sin?The Derivative 83 46 y= fever 4122 y = Infe®) 4417 y = xsin x sinh x 4123 y = sous 418 y = S/[(L + tanh x}?] 414 y= JF — BR) atid y= (PL +e) 4125 y = sins 41200, y = xe 4126 y = tan*3x) 4121 y = sect tan x 4127 y = JsinGx — 4) In Problems 4-128 through 4-139 find the second derivative: 418 f(x) = x8 — 3x9 + 2x 1 4134 f(x) = tan x 4129 f(x) = x9 2x! 4135 f(x) = 2° 4130 y= Je + (Ys) 4-136 f(x) = arc tan x 4131 y= Jr? - x 4137 y = xe" 4132 y =e! 4138 y = In(l + 2x) 4133 y= (x + yx — 1) 4139 y = In Problems 4-140 through 4-150 find the fourth derivative: 4140 f(s) Saxte eget 4-146 f(x) = In(cos x) 4141 f(x) = 36 4147 f(x) =e 4142 f(x) = Inx 4148 f(x) = PS 4143 f(x) = sin 3x + cos 2x 4149 flx) = xaretan x — fIn(l + 32) 414 fox) = 4-9) 4150 f(x) = 5x" 4145 f(x) = cosh(I — 2x) Solutions to Supplementary Exercises im (508 + Ax) +4] ~ (50 +4) (438) 0) = Jim, x (439) 0) = im, Oe a ate ee oo aa ax +) gy ao (435) pt = tin PATE AT HDs ane + fet Ort84 Caleulus (639 £9) = tin VRE EB 1 VET Ux + Ax) — 1 2x — 1) ain AxLfile+ Ax) 1+ J2e— 1) JT eave et a wd rs (437) f'(0) = is = 3x + 2e (3) 0 = ti, OEE AE Ml + BA) — + HU) +0 Ax (+ Ax) oii + ye Fay ea ne 9 x (429 70) tin EM 2 aa) = We = tm [ xt Ax — x © geo Laxi/x + Ax + J) [ ar + verari]-[, : (4-40) fe) = Jim Le 7 axe Wet Ox) * Ane bae—1+ 3 =F fs == sie Ax= 0) | (Fant (441) £09 = lim +ayet] G+ 1+ ax) (444) f(x) = 10x — 13 (4.45) y' = dic + 6x? + 10x +6 a y = (8150 (4-47) y' = (5/6) N° (4-48) y' = Sixt (4-49) f"(0) = 2x + (In 232" (4:80) f(x) = 30s + 4 see? x (451) y= sinx + x08 x (452) y\ = xe" + Pet 4 Se — Bre (453) y = 0—axifts — 1 = assy = —Vlll ~ YP] (456 9" = nx — Dilla?) (457) £09) = WML) fT 34) ~ are sin x] In 2x in? 3) in x)/{(1 + c0s x] + £08 xoas x ~ sin x(o0s x ~ 8 {sine + co x)? © (gin x + 08 x)? (4-62) y= (463) fa) = BU +” (4-64) f(x) = 2x + Yr 4? (465) fa) = B97 + He bye (4-66) f(x) = fsin” "x c08 x (467) y' = 2x/(x? + 2) (4-68) y' = 3 tan?x sect (469) y= [Ut + ae (470) y= A bet et (41) y = 4G3 + x — 99 Gx? + 1) (4472) y* = sinb(sinh xjoosh x (473) »" (474) (9) = in (cos x)25* (4-75) f°(x) = (In 2)2* cos 2* (4-76) fx) = (In sin 2sin"2 (HTT) f(x) = = 10x/[(x? — 2°] (4678) = Hsin x + cos x)-#2(e08 x ~ sin x) (419) y= [In 2/2)]2"* (4-80) y' = fese%(1/a)Ie"?) (4-81) yf! = —2(? — 4x7 + 3) [Dx = (28/3)x4°] (689) y= xii = 1) (4-83) y' = cos(xe*Ye* + xe*) (4684) y' = Hsin x!) "eos '2)hx “1 = (00s xylan) The Derivative 8586 Calculus (485) f°(x) = 2x In sin x + x2((608 x)(6in x] (4-86) f(x) = F/T + PT (487) FG) = Bx? = (5/2) 1F = (7/2) (4-88) y° = (cos xfer) (4-89) y= -231(x + Nee ~ 5° (4.90) yf = (+ AE — Die + DTN = W/E + OP (491) $00) = [=x ~ J? — IB = VF 1) (492) f(s) = (esin T= VT — #7) (493) f'0) = 2x(Qx? ~ 11) + 6202 — 3) = 1x4 ~ 18? ~ 22x 494) (9 = Ce + JT + YTS = VT (4-95) f(s) = Billet + oF] (4-96) y" = (4 In@x ~ DI2x ~ 1) (4-97) y’ = (L + etfs + e*)Intx + eI] . (498) y= ot + 2ey/LIn S60" + 71) (4-99) yf = 10 In 10Jogyo x + [10%/¢e In 10)} (4-100) yf = Sx(x? — 17% + HQ 4 fx? = AL + xty Mayo (4101) y! = Sflsin x)x}*[ 60s x — sin 9/3] (4-102) y/ = (in x(x + 1) + [Inte + EY) (4-103) y= (U3) + [1/0 + Dnt + 1) (4-104) y’ = [3 tan*(In x) soe? (In x)]/xx (4-105) y' = -e* ewe cot e* — ese x cot xe (4106) y’ = 31n¥Q + sin [00s 0/2 + sin xt] (4107) 9° = Gs) + [00s x)f6in x] ~ [2x/lo? + 19) (4-108) y” (4-109) "= ext fe (4-110) f"(8) = cos x cos(sin x) + 2sin x cos x (411) fo) = 0 (4-112) y' = ~Ssech’3 - 5x) (4-113) y’ = flogy " x(1/In 3)(1/x) = 1/(2x /log, x In 3) etd) f) = He + 1! e — IN = (1/3 = ER PIMC + PP (4-115) f"(a) = (S/6K1 + sin x97"? sin x 008 (atte) y' = Aye + Dix + $y xibe + 1)a 1 (4-118) y’ = (—10 sech?x)/[(1 + tanh x)*] (4119) y’ = Qe — eFC + 7] (4-120) ¥ = 4e7E1 +) (4121) yf = Asects tants + sec (#12) y= 3 (4-123) y= In Ssinh x Se (4-124) y= Het — 3x4 er 4 3x74) (4-125) y= 0 (4-126) ¥ = 6tan 3x sec*3x (4-127) y= [3 cos(3x — 4)]/[2,/sinfx — 4)) (4-128) f(x) = 20%? = 18x (4-129) f"(x) = 6x ~ 4x-* (4-130) 9” =p? gta (4131) y" = 08/9) Ie (4-132) y" = 160"! (4-133) y" = 4/[ — 07 (4-134) /"x) = 2see? x tan x ‘The Derivative 87 (4117) y= sin xsinh x + x 0s x sinh x + x sin x cosh x (4135) $3) = (In? 22" (4136) x) = —2x/{C1 + 27] (4137) y" = 20" + xe* (4138) y" = ~4/f(L + 29°) (4139) 9" = 2010 3)3" + 4x21? 393° (4-140) f(x) = 120% — 24 (4141) f(x) = 48e™* (4142) f°) = ~6x* (4143) f(x) = 81 sin 3x + 16 608 2x (144) 4) = 24(1 — x)? (4145) f(x) = 16 cosh(l ~ 2x) ($146) f(x) = 4 sec?e tan?x — 2 sects (4-147) f(x) = de"(3 + 12x? + 4x4) (4-148) f(x) = 16 (n* 39375 (4149) fa) = (6x? — QTL + x7P] (4-150) (x) = (—400/81)x-4| __EXAM 1 (CHAPTERS 1-4) i 1. Find the equation ofthe line through (1,2) perpendicular to theline through (2, S)and (3, ~4) i in slope-intercept form. | 2, Find the partial fraction decomposition of xe 3, Use the identity for cos 26 to find cos(2 arc tan (1/4)) 4, Identify, put in standard form, and graph x? = dy? - T=0 | 5, Find the following limits: x= (Us) sin x 5 er rr Opes eet 6. Use the definition to find the derivative of f(x) = 1/(« + 1) 7, Determine whether f(x) has any discontinuities: 3 So) =43 12. Qtr) : 13. w= Ing + 14, he) = see 10. f(x) = 02 +x 4) / fo 4x8 1s ¢= at (!) 11. G(x) = sin?x \x SOLUTIONS TO EXAM 1 1. The line joining (2, 5) and (3, ~4) has slope 5 m= Aine perpendicular to this line has slope 1/9. Thus you want the line a 19: y ) 1 ot, ye yagrgt2agttExam 1 89 aoe ert x BL Cx+D caren Clear the denominators: 37 2 = Axle? + 1) + Ble? +1) + (Cx + D)x? = x4 + C)+ (B+ D) 4 xAtB and equate the coefficients: O=44C 1=B4D O=4 2=B Solve using A = 0,B = -2,C = 0,D =3: x2 cat 3 ware x? xt ET 3, The identity for cos 20s cos 20 = 2.cos? 0 — 1, Let = arctan(t/4}. Draw ‘ a triangle with an angle 6, as shown in the figure. Then tan @ = (1/4) and cos @ = 4/,/17. Finally, vi ; L : 15 cos| (2 arc tan 5) = cos(28) = 2cos*O i 4 Because the coeficients of x? and y? are opposite in sign, the graph is hyperbola, Put the equation in standard form by completing the square in x and y: | xP sh -dy? 4 2yt =741-4 | (x= 1 = 4(y + IP =4 The center is (1, —D), with a= 2 b= 1. The asymptotes are y + 1 = + (1/2)(x — 1), as in the figure. | =) Pt = e+) 5) ee i Se bial 4 x 1 sig Spy SX B= ons 4 sin. x(1 — 005») T= cos*x. F008 x ere T+ COS x sinx(1_— e083) _ 4, 1 — eos x were SIN. lim wn 13K90° Calculus 1 1 6. tim LE +49) = 0) x+ dx am x+1 (+ N-(@+ Ax +1) Qe + Doe + Ax + Dax im ———~4*__ seo (x + I + Ax + 1)Ax | | | -t -1 = I et ee Ox ea P 7. The only possible discontinuities are at x = 1 and x = 5. Look at the one-sided limits at x= Land x = 5: lim fe) wn (5x43) feist lim f(x) = lim 2 = 2 Because lim,.,- f(x) # lim,.,+ f(x), the limit lim,.., f(x) does not exist and f(x) is not continuous atx = 1 lim f(x) = lim 2 = ass ms 4 lim fx) = lim (7 ~ x) = 2 asst a ‘Thus, lim,..5- f(x) = limy..s+ f(x) = f(5), and so f(x) is continuous at x = 8. Use the quotient rule F & 9408 + 3-04 dot - 3 Ze HF ‘ Ge? ~ 5) — (x + 3)Qx) OF 5F | 34) Ba 5 3) = 23 (=! 61 2 as n+ JS (203) = 2¢ ()* /5 — + (61?) 10, Use the chain rule: SQ) = 5x0? + x — athe $x- 4 = 50? +x — 42x +1) IL G(x) = 3 sin? x ¢08 x 12, ayy ey (i= 5 7 Jae = ay (ay =1.4 a jen) - 7 Bi geaet gly afte+ JA Wo) = see fi tan Je 4 (ya = SEV BOE dx aye dv a et aly ot fet) “T+ Gaye de \x) ~ 14+ OP) PFT Exam} 91