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Unformatted text preview: ∗ 7.9 THE OTHER HYPERBOLIC FUNCTIONS 443 as the only possibility. Taking the natural log of both sides, we get y = ln (x + The proof of (iii) is left as an exercise. EXERCISES 7.9
Differentiate the function. = tanh x. = ln ( tanh x). = sinh ( tan−1 e2x ). √ = coth ( x2 + 1). sech x 9. y = . 1 + cosh x 1. 3. 5. 7. y y y y
2 x2 − 1). 21. Show that 2. 4. 6. 8. y = tanh 3x. y = tanh (ln x). y = sech (3x2 + 1). y = ln (sech x).
2 (7.9.5) d 1 ( tanh−1 x) = , dx 1 − x2 −1 < x < 1. 22. Show that d −1 (sech−1 x) = √ , dx x 1 − x2 23. Show that d −1 (csch−1 x) = √ , dx |x| 1 + x2 24. Show that d 1 , ( coth−1 x) = dx 1 − x2 |x| > 1. x = 0. 0 < x < 1. 10. y = cosh x . 1 + sech x Verify the differentiation formula. 11. 12. 13. 14. d ( coth x) = − csch2 x. dx d (sech x) = − sech x tanh x. dx d (csch x) = − csch x coth x. dx Show that tanh (t + s) = tanh t + tanh s . 1 + tanh t tanh s 15. Given that tanh x0 = 4 , ﬁnd (a) sech x0 . 5 HINT: 1 − tanh2 x = sech2 x. Then ﬁnd (b) cosh x0 , (c) sinh x0 , (d) coth x0 , (e) csch x0 . 5 16. Given that tanh t0 = − 12 , evaluate the remaining hyperbolic functions at t0 . √ 17. Show that, if x2 ≥ 1, then x − x2 − 1 ≤ 1. 18. Show that tanh−1 x = 19. Show that
(7.9.3) 1 1+x ln , 2 1−x −1 < x < 1. 25. Sketch the graph of y = sech x, giving: (a) the extreme values; (b) the points of inﬂection; and (c) the concavity. 26. Sketch the graphs of (a) y = coth x, (b) y = csch x. 27. Graph y = sinh x and y = sinh−1 x in the same coordinate system. Find all points of inﬂection. 28. Sketch the graphs of (a) y = cosh−1 x, (b) y = tanh−1 x. 29. Given that tan φ = sinh x, show that dφ = sech x. (a) dx (b) x = ln ( sec φ + tan φ ). dx (c) = sec φ . dφ 30. The region bounded by the graph of y = sech x between x = −1 and x = 1 is revolved about the x-axis. Find the volume of the solid that is generated. Calculate the integral. 31. tanh x dx. sech x dx. sech3 x tanh x dx. 32. 34. 36. coth x dx. csch x dx. x sech2 x2 dx. d 1 ( sinh−1 x) = √ , 2+1 dx x x real. 20. Show that 33.
(7.9.4) d 1 ( cosh−1 x) = √ , dx x2 − 1 x > 1. 35. 444
37. 39. CHAPTER 7 THE TRANSCENDENTAL FUNCTIONS tanh x ln ( cosh x) dx. sech2 x dx. 1 + tanh x 38. 40. 1 + tanh x dx. cosh2 x tanh x sech x dx.
5 2 (a) Show that v (t ) = mg tanh k gk t m In Exercises 41–43, verify the given integration formula. In each case, take a > 0. 41. 42. √ √ 1 a2 + x2 1 x 2 − a2 dx = sinh−1 dx = cosh
−1 x + C. a x + C. a is a solution of the equation which satisﬁes v(0) = 0. (b) Find lim v(t ).
t →∞ This limit is called the terminal velocity of the body. c 45. Use a CAS to ﬁnd a one-term expression for f .
(a) (b) f (x) dx = − f (x) dx = sech x(2 + tanh x) . 3(1 + tanh x)2 1 tanh−1 x + C If |x| < a. a 1 a 43. dx = 1 a2 − x 2 coth−1 x + C If |x| > a. a a 44. If a body of mass m falling from rest under the action of gravity encounters air resistance that is proportional to the square of its velocity, then the velocity v(t ) of the body at time t satisﬁes the equation m dv = mg − kv2 dt ln (x4 − 1) x2 tanh−1 x2 + . 2 4 Verify your results by differentiating the right-hand sides. c 46. Use a CAS to ﬁnd a one-term expression for f . x x (a) f (x) dx = 16 + x2 − 8 sinh−1 . 2 4
x2 + 1 x2 −1 x2 cosh−1 x2 + . 2 x2 + 1 2 Verify your results by differentiating the right-hand sides. (b) f (x) dx = − where k > 0 is the constant of proportionality and g is the gravitational constant. CHAPTER HIGHLIGHTS
7.1 One-to-One Function; Inverses cot x dx = ln | sin x| + C , csc x dx = ln |csc x − cot x| + C . logarithmic differentiation (p. 394)
7.4 The Exponential Function one-to-one function; inverse function (p. 371) one-to-one and increasing/decreasing functions; derivatives (p. 374) relation between graph of f and the graph of f −1 (p. 375) continuity and differentiability of inverse functions (p. 375) derivative of an inverse (p. 376)
7.2 The Logarithm Function, Part 1 deﬁnition of a logarithm function (p. 381) x dt , x > 0; natural logarithm: In x = t t domain (0, ∞), range (− ∞, ∞) basic properties of ln x (p. 385) graph of y = ln x (p. 385)
7.3 The Logarithm Function, Part II The exponential function y = e x is the inverse of the logarithm function y = ln x. graph of y = e x (p. 398); domain (− ∞, ∞), range (0, ∞) basic properties of the exponential function (p. 398) du du (e ) = e u , dx dx
7.5 eg (x) g (x) dx = eg (x) + C Arbitrary Powers; Other Bases d 1 du (ln |u|) = , dx u dx g (x ) dx = ln |g (x)| + C , g (x ) tan x dx = ln | sec x| + C , sec x dx = ln | sec x + tan x| + C , xr = er ln x for all x > 0, all real r logp x = ln x ln p du du (p ) = pu ln p (p a positive constant), dx dx d 1 du ( logp u) = dx u ln p dx 1+ 1 n
n ≤e ≤ 1+ 1 n n+1 e ∼ 2. 71828 = ...
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This note was uploaded on 10/12/2010 for the course MATH 12345 taught by Professor Smith during the Spring '10 term at University of Houston - Downtown.
- Spring '10
- Hyperbolic Functions