## DLS 2 years ago y=sinx^cosx+cosx^sinx dy/dx=?

1. DLS

My attempt:

2. DLS

$\LARGE \frac{1}{y} \times \frac{dy}{dx} =\frac{cosx}{sinx} \times (-sinx) + (cosx)(-cosec^{2}x) + \frac{sinx}{cosx}$ $\LARGE \times (cosx)+(sinx)(\sec^{2}x)$

3. DLS

last term hidden is sinx/cos then continued below

4. hartnn

what was that ?!! y= u+v find du/dx and dv/dx using logarithmic differentiation then add...

5. DLS

lol

6. DLS

I used chain rule for logsinx then UV rule with (cosx)logsinx

7. DLS

$\LARGE logy=(cosx)(logsinx)+(sinx)(logcosx)$

8. hartnn

log (A+B) cannot be simplified to log A +log B

9. DLS

where is log(A+B) :O

10. DLS

oh

11. DLS

NEVERMIND

12. hartnn

y= u+v find du/dx and dv/dx using logarithmic differentiation then add... thats the only way...

13. abb0t

Use chain rule.

14. DLS

$\LARGE logy=\log(cosx^{cosx} \times -sinx + (-sinx^{sinx} \times cosx)$

15. DLS

?

16. hartnn

u =sin x^cos x du/dx =... ?

17. DLS

again log? O_O

18. DLS

okay no

19. hartnn

again ? this is the 1st time, you use log...

20. DLS

:/

21. DLS

cosxlogsinx?

22. DLS

CONFUSED

23. hartnn

log u = cos x log sin x now diff. use product rule.

24. DLS

thats what I did.. I got: cotx cosx

25. hartnn

did u forget product rule ?

26. DLS

:/ no :///

27. hartnn

(uv)' = uv'+u'v

28. DLS

$cosxlogsinx= cosx(\frac{1}{sinx}) + logsinx (-sinx)$

29. DLS

isnt it :/..

30. hartnn

1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)

31. DLS

I guess I studied too much :P

32. DLS

thats what I did >.<

33. hartnn

extra cos x come due to chain rule is it ?

34. DLS

cosxlogsinx =>cosx(1/sinx * cosx)(chain rule ) cosxcotx now product rule whats wrong with it :/

35. hartnn

because there was no 2nd term in there....thats why i asked u about product rule...

36. DLS

stil confused :|

37. hartnn

ok, just tell me what u get finallu for du/dx .. ?

38. hartnn

u get this, right ? 1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)

39. DLS

let me write

40. hartnn

don't write in latex, takes time unnecessarily...

41. DLS

yes,im solving last time :o only du/dx

42. DLS

i get

43. hartnn

yes, only du/dx....dv/dx will be very similar....

44. DLS

-cosx cosec^2 x -sinxcotx :///////

45. DLS

on solving cosxcotx by product rule thats it

46. hartnn

:( didn't u get this ? 1/u du/dx = cos x (1/ sin x) (cos x) - sin x log (sinx)

47. DLS

I dont think so loll :/

48. hartnn

log u = cos x log sin x this ?

49. DLS

yes

50. DLS

cos x (1/sinx * cosx)

51. DLS

log dissapeared

52. hartnn

(log u)' = (cos x log sin x )' 1/u u' = cos x (log sin x)' + log sin x (cos x)' did u get this step ?

53. hartnn

just simple product rule.

54. DLS

okay

55. hartnn

want to continue ? or should i ?

56. DLS

i will

57. DLS

logsinx(-sinx)+cosx(logsinx) okay?

58. DLS

DUH

59. hartnn

my next step 1/u u' = cos x (log sin x)' + log sin x (cos x)' 1/u u' = cos x (1/ sin x)(sin x)' + log sin x (-sin x) get that ? ^

60. hartnn

use of chain rule.

61. DLS

1minute and its over!

62. DLS

give me

63. DLS

lol been 30 minutes:/

64. hartnn

let me do it...

65. hartnn

u just see.

66. DLS

ivedone

67. DLS

$\LARGE logsinx(-sinx)+cosx(\frac{1}{sinx} \times cosx)$ RIGHT?

68. hartnn

1/u u' = cos x (log sin x)' + log sin x (cos x)' 1/u u' = cos x (1/ sin x)(sin x)' + log sin x (-sin x) 1/u u' = cos x (1/ sin x)(cos x) - sin x log sin x du/dx = u [cos x . cot x - sin x log sin x] finally du/dx = [sin x^cos x][cos x . cot x - sin x log sin x] any doubts in any step ?

69. DLS

$FINAL:logsinx(-sinx)+cosxcot$

70. DLS

x

71. DLS

this is what u get out of cosx log sinx right? ^ and im doing it myself...ill check from urs,this is very confusingnow just tell me that^ though i checked with WF

72. hartnn

du/dx = [sin x^cos x][cos x . cot x - sin x log sin x] in same manner find dv/dx.

73. hartnn

74. DLS

YES! correct!

75. DLS

that was one hell of a question -_-'

76. hartnn

but easy.

77. DLS

for u!

78. hartnn

sum of 2 eaSY Q's

79. DLS

it was new for me..so yeah..next time easy for me too...(hopefully)