CHAPTER 25 Derivatives Of Inverse Trig Functions

CHAPTER 25Derivatives of Inverse Trig FunctionsOur goal is simple, and the answers will come quickly. We will derive sixnew derivative formulas for the six inverse trigonometric functions:d h 1 isin (x)dxd h 1 icos (x)dxid htan 1 (x)dxd h 1 icot (x)dxd h 1 isec (x)dxd h 1 icsc (x)dxThese formulas will flow from the inverse rule from Chapter 24 (page 278):d h 1 i1f (x) 0 1 .dxf f (x)(25.1)25.1 Derivatives of Inverse Sine and CosineApplying the inverse rule (25.1) with f (x) sin(x) yieldsd h 1 i1sin (x) 1 .dxcos sin (x)(25.2)We are almost there. We just have to simplify the cos sin 1 (x) in the denominator. To do this recallsin 1(x) the angle º2 µ º2for which sin(µ ) x!.Thus sin 1 (x) It is the angle (between º2 and º2 ) of a thetriangle on the unit circle whose opposite side is x. (Be cause sin of this angle equals x.) Then cos sin 1 (x) is thelength of the adjacenttheoremp side. By the Pythagorean 1 p2this side length is 1 x . Putting cos sin (x) 1 x2into the above Equation (25.2), we get or latest rule:Rule 20d h 1 i1sin (x) pdx1 x21sin 1 ( x) {z} cos sin 1 (x)x

287Derivatives of Inverse Sine and CosineWe reviewed sin 1 (x) In Section 6.1 and presented its graph on page 101.Figure 25.1 repeats the graph, along with the derivative from Rule 20.1f 0 (x) p1 x2yf (x) cos 1 (x)f (x) sin 1 (x)º21-11x º2Figure 25.1. The function sin 1 (x) and its derivative. The derivativeisalways positive, reflecting the fact that the tangents to sin 1 (x) have positiveslope. The derivative has vertical asymptotes at x 1, as the tangents tosin 1 (x) become increasingly steep as x approaches 1.Now consider cos 1 (x). The tangents to its graph (Figure 25.2 below)have negative slope, and the geometry suggests that its derivative is negativethe derivative of sin 1 (x). Indeed this turns out to be exactly the case. Thischapter’s Exercise 1 asks you to prove our next rule:Rule 21d h 1 i 1cos (x) pdx1 x2f (x) cos 1 (x)-1º 11x 1f 0 (x) p1 x2Figure 25.2. The function cos 1 (x) and its derivative.

288Derivatives of Inverse Trig Functions25.2 Derivatives of Inverse Tangent and CotangentNow let’s find the derivative of tan 1 (x). Putting f (x) tan(x) into the inverserule (25.1), we have f 1 (x) tan 1 (x) and f 0 (x) sec2 (x), and we getid h11 1 tan (x) 2 . 12dxsec tan (x)sec tan 1 (x) The expression sec tan 1 (x) in the denominator is thelength of the hypotenuse of the triangle to the right.(See example 6.3 in Chapter 6, page 114.) By the Pythagoreantheorem, the length is sec tan 1 (x) p1 x2 . Inserting this into the above Equation (25.4)yields 1 antcesRule 22 x( )tan 1 ( x)x1id h111tan 1 (x) . 2 2 p 12dx1 x21 xsec tan (x)We now have:(25.3)id h1tan 1 (x) dx1 x2We discussed tan 1 (x) in Chapter 6, and its graph is in Figure 6.3. BelowFigure 25.3 repeats the graph, along with the derivative x21 1 .f (x) tan 1 (x)º211f (x) 1 x2x0 º2Figure 25.3. The function tan 1 (x) and its derivative1.1 x 2Note lim12 0x!1 1 x 1and lim 1 1x2 0, reflecting the fact that the tangent lines to y tan (x)x! 1become closer and closer to horizontal as x ! 1. The derivative bumps upto 1 at x 0, where the tangent to y tan 1 (x) is steepest, with slope 1Exercise 3 below asks you to mirror the above arguments to deduce:Rule 23d h 1 i 1cot (x) dx1 x2

289Derivatives of Inverse Secant and Cosecant25.3 Derivatives of Inverse Secant and CosecantWe reviewed sec 1 (x) in Section 6.3. For its derivative, put f (x) sec(x) intothe inverse rule (25.1), with f 1 (x) sec 1 (x) and f 0 (x) sec(x) tan(x). We getd h 1 i1sec (x) 0 1 dxf f (x)1 sec f 1 (x) · tan f 1 (x)1 . 1sec sec (x) · tan sec 1 (x) Because sec sec 1 (x) x, the above becomesd h 1 i1 .sec (x) dxx · tan sec 1 (x) In Example 6.5 we showed that tan sec 1 (x) With this, Equation 25.4 above becomes8 1phidx x2 1sec 1 (x) 1 dx :p x x2 1((25.4)px2 1p x2 1if x is positiveif x is negativeif x is positiveif x is negative.But if x is negative, then x is positive, and the above consolidates toRule 24d h 1 i1sec (x) pdx x x2 1This graph of sec 1 (x) and its derivative is shown in Figure 25.3.yºy f ( x) sec 1 ( x)º2 11y f 0 (x) p1 x x2 1Figure 25.4. The graph of sec 1 (x) and its derivative. The domain ofboth functions is ( 1, 1] [ [1, 1). Note that the derivative has verticalasymptotes at x 1, where the tangent line to y sec 1 (x) is vertical.

290Derivatives of Inverse Trig FunctionsThis chapter’s Exercise 2 asks you to use reasoning similar to the aboveto deduce our final rule.Rule 25d h 1 i 1csc (x) pdx x x2 1Each of our new rules has a chain rule generalization. For example,Rule 25 generalizes as id h 1 1 g0 (x)0cscg(x) Øg(x) .qqØØØdxØ g(x)Ø (g(x))2 1Ø g(x)Ø (g(x))2 1Here is a summary of this Chapter’s new rules, along with their chain rulegeneralizations.d h 1 isin (x)dxd h 1 icos (x)dxid htan 1 (x)dxd h 1 icot (x)dxd h 1 isec (x)dxd h 1 icsc (x)dxExample 25.1Example 25.2Example 25.3 1p1 x2 1p1 x211 x2 11 x2 p 1 x x2 1 1p x x2 1 id h 1 sing(x)dx id h 1 cosg(x)dx id htan 1 g(x)dx id h 1 cotg(x)dx id h 1 secg(x)dx id h 1 cscg(x)dx qq121 (g(x)) 1g0 (x)g0 (x)1 (g(x))21g0 (x)21 (g(x)) 11 (g(x))2g0 (x)g0 (x)qØØØ g(x)Ø (g(x))2 1 g0 (x)ØØqØ g(x)Ø (g(x))2 1d hp 1 id h 1 12 i 1 1 12 d h 1 icos (x) cos (x) cos (x)cos (x)dxdx2dx 1 11 1 cos 1 (x) 2 p p.p2 121 x2 cos (x) 1 x2 1i 1 1d h tan 1 ( x) id h1etan ( x)e etan ( x)tan 1 (x) etan ( x) .dxdx1 x21 x 2 id h1d h xi1exxtan 1 e x e e .dx1 e2 x1 e2 x1 (e x )2 dx

291Summary of Derivative Rules25.4 Summary of Derivative RulesWe have reached the end of our derivative rules. In summary, we havethe following rules for specific functions. The corresponding chain rulegeneralizations are shown to the sLogRulesTrigRulesInverseTrigRulesd §c 0dxd §x 1dxd h nix nx n 1dxd h xie exdxd h xia ln(a)a xdx§ 1d hln( x) dxxid h1loga ( x) dxx ln(a)id hsin( x) cos( x)dxid hcos( x) sin( x)dxid htan( x) sec2 ( x)dxid hcot( x) csc2 ( x)dxid hsec( x) sec( x) tan( x)dxid hcsc( x) csc( x) cot( x)dxd h 1 i1sin ( x) pdx1 x2hid 1cos 1 ( x) pdx1 x2hid1tan 1 ( x) dx1 x2d h 1 i 1cot ( x) dx1 x2hid1sec 1 ( x) pdx x x2 1hid 1csc 1 ( x) pdx x x2 1Chain Rule Generalization n § n 1 0d g ( x) n g ( x)g ( x)dxd h g ( x) ie e g ( x) g 0 ( x )dxd h g ( x) ia ln(a)a g( x) g0 ( x)dx id h 1 0ln g( x) g ( x)dxg ( x) id h1g 0 ( x)loga g( x) dxg( x) ln(a) i d hsin g( x) cos g( x) g0 ( x)dx i d hcos g( x) sin g( x) g0 ( x)dx i d htan g( x) sec2 g( x) g0 ( x)dx i d hcot g( x) csc2 g( x) g0 ( x)dx i d hsec g( x) sec g( x) tan g( x) g0 ( x)dx i d hcsc g( x) csc g( x) cot g( x) g0 ( x)dx id h 1 1sing ( x) pg 0 ( x)2dx1 ( g( x)) id h 1 1cosg ( x) pg 0 ( x)2dx1 ( g( x)) id h1 1tang ( x) g 0 ( x)dx1 ( g( x))2 id h 1 1cotg ( x) g 0 ( x)dx1 ( g( x))2 id h 1 1secg ( x) g 0 ( x)pdx g( x) ( g( x))2 1 id h 1 1cscg ( x) g 0 ( x)pdx g( x) ( g( x))2 1

292Derivatives of Inverse Trig FunctionsIn addition we have the following general rules for the derivatives ofcombinations of functions.§d c f (x)dx§d f (x) g(x)dx§d f (x)g(x)dx d f (x)dx g(x) §d f g(x)dxd 1 §f (x)dxConstant Multiple Rule:Sum/Difference Rule:Product Rule:Quotient Rule:Chain Rule:Inverse Rule: c f 0 (x) f 0 (x) g0 (x) f 0 (x)g(x) f (x)g0 (x) f 0 (x)g(x) f (x)g0 (x)(g(x))2 f 0 (g(x)) g0 (x) f0 1f 1 (x) We used this last rule, the inverse rule, to find the derivatives of ln(x)and the inverse trig functions. After it has served these purposes it is mostlyretired for the remainder of Calculus I, except for the stray exercise or quizor test question.This looks like a lot of rules to remember, and it is. But through practiceand usage you will reach the point of using them automatically, with hardlya thought. Be sure to get enough practice!Exercises for Chapter 25d h 1 i 1cos (x) p.dx1 x2d h 1 i 12. Show thatcsc (x) p.dx x x2 1d h 1 i 13. Show thatcot (x) .dx1 x21. Show thatFind the derivatives of the given functions. p 2x 7. tan 1 º x 10. cos 1 º x 13. tan 1 ln(x) º4. sin 1 5. ln tan 1 (x)6. e x tan 1 (x)8. sec 1 º x9. ln sin 1 (x) 11. sec 1 x5 14. tan 1 x sin(x) 112. etan(º x) 15. x sin 1 ln(x)

293Summary of Derivative RulesExercise Solutions for Chapter 25d h 1 i 1cos (x) p.dx1 x2d h 1 i1 .By the inverse rule,cos (x) dx sin cos 1 (x)1. Show thatNow we simplify the denominator.From the standard diagram for OPPcos 1 (x) we get sin cos 1 (x) HYP p1 x2 . With this, the aboved h 1 i 1becomescos (x) p.dx1 x2p1 x 21 1cos 1 (x)ºd h 1 i 1cot (x) .dx1 x2Suggestion: Verify the identity cot 1 (x) 0x3. Show thatboth sides of this.5.º2 tan 1 (x). Then differentiatei id h 1d h111 1 ln tan 1 (x) tan(x) dxtan 1 (x) dxtan 1 (x) 1 x2 tan 1 (x) 1 x2 id hººtan 1 º x 2dx1 (º x)1 º2 x 2d h 1 i1d h 1 i1119.ln sin (x) sin (x) pp 1 1 12dxsin (x) dxsin (x) 1 xsin (x) 1 x27.11.13.15.d h 1 5 i15x45secx Ø Ø q 5x4 Ø Øp Ø ØpØ x5 Ø x10 1 Ø xØ x10 1dxØ x5 Ø x5 2 1i d h111tan 1 ln(x) º 2 2dx1 ln(x) x x x ln(x) i id hd h 1 x sin 1 ln(x) 1 · sin 1 ln(x) xsin ln(x)dxdx 111 1 sin 1 ln(x) x qln(x) q 2 x sin 21 ln(x)1 ln(x)

288 Derivatives of Inverse Trig Functions 25.2 Derivatives of Inverse Tangent and Cotangent Now let’s find the derivative of tan 1 ( x).Putting f tan(into the inverse rule (25.1), we have f 1 (x) tan and 0 sec2, and we get d dx h tan 1(x) i 1 sec2