In Exercises 21–42, find the derivative of y with respect to the appro- priate variable. 21. y = cos(x²) 23. y = sin-lV2t 22. y = cos(1/x) 24. y = sin(1 – t) 25. y = sec-l(2s + 1) 26. y = sec-15s 27. y = csc-(x² + 1), x> 0 28. y = csc-1 sec-, 0 <1< 1 29. y = 30. y = sin 31. y = cot Vi 33. y = In (tanlx) 32. y = cotl Vi – 1 34. y = tan¬ (In x) 35. y = csc-'(e') 36. y = cos¯'(e¯") 37. y = sV1 – s² + cos¯ls 39. y = tan¬l vx² – 1 + csc-1x, 38. y = Vs² - 1 sec-ls x > 1 40. y = cot 41. y = xsinx + Vĩ – x² - tan¬lx 42. y = In (x² + 4) – x tan IN MI2

Calculus: Early Transcendentals
8th Edition
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Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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In Exercises 21–42, find the derivative of y with respect to the appro-
priate variable.
21. y = cos(x²)
23. y = sin-lV2t
22. y = cos(1/x)
24. y = sin(1 – t)
25. y = sec-l(2s + 1)
26. y = sec-15s
27. y = csc-(x² + 1), x> 0 28. y = csc-1
sec-, 0 <1< 1
29. y =
30. y = sin
31. y = cot Vi
33. y = In (tanlx)
32. y = cotl Vi – 1
34. y = tan¬ (In x)
35. y = csc-'(e')
36. y = cos¯'(e¯")
37. y = sV1 – s² + cos¯ls
39. y = tan¬l vx² – 1 + csc-1x,
38. y = Vs² - 1
sec-ls
x > 1
40. y = cot
41. y = xsinx + Vĩ – x²
- tan¬lx
42. y = In (x² + 4) – x tan
IN MI2
Transcribed Image Text:In Exercises 21–42, find the derivative of y with respect to the appro- priate variable. 21. y = cos(x²) 23. y = sin-lV2t 22. y = cos(1/x) 24. y = sin(1 – t) 25. y = sec-l(2s + 1) 26. y = sec-15s 27. y = csc-(x² + 1), x> 0 28. y = csc-1 sec-, 0 <1< 1 29. y = 30. y = sin 31. y = cot Vi 33. y = In (tanlx) 32. y = cotl Vi – 1 34. y = tan¬ (In x) 35. y = csc-'(e') 36. y = cos¯'(e¯") 37. y = sV1 – s² + cos¯ls 39. y = tan¬l vx² – 1 + csc-1x, 38. y = Vs² - 1 sec-ls x > 1 40. y = cot 41. y = xsinx + Vĩ – x² - tan¬lx 42. y = In (x² + 4) – x tan IN MI2
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