## anonymous 3 years ago Please anyone help me with this question. Find Y prime: Y=sin^-1(2x^3).

1. anonymous

When you say y prime, do you mean the derivative?

2. anonymous

yup

3. anonymous

Please someone help me I m stuck and its due tomorrow

4. goformit100

Yo Yo, diffrentiate on both sides...

5. anonymous

These are kind of new problems to me if you dont mind can you teach me and help me solving them?

6. anonymous

7. anonymous

can we go step by step?

8. anonymous

Ya sure, I'll set it up.

9. anonymous

you want to go by steps sure will do

10. anonymous

at the end i will show u how u can work such equation in ur head, without paper and pen

11. anonymous

Thanks to both of you Sujay and Mathsmind :)

12. anonymous

Original Equation:$y=\sin^2-1(2x-3)?$

13. anonymous

with an x after the sin

14. anonymous

I want to make sure my original is correct

15. anonymous

$y=\sin^{-1}(2x^3)$

16. anonymous

17. anonymous

are u there?

18. anonymous

ok let me write down the equation here. Its $y=\sin^{-1} (2x ^{3})$

19. anonymous

i did write it

20. anonymous

yeah I m here sorry I was just writing that equation down

21. anonymous

yeah you are right Mathmind

22. anonymous

ok there are 2 main ways to solve this problem at ur level

23. anonymous

is there any easy way bcoz I have 4 more to do after this one so I need to understand the whole thing my professor method was too hard

24. anonymous

$y = \arcsin(x)$

25. anonymous

$\sin(y)=x$

26. anonymous

differentiate both sides

27. anonymous

Original:$\sin^{-1} (2x^3)$ Use the chain rule: Derivative out the outside times the derivative of the inside. $-\sin(2x^3)[\cos(2x^3)](6x^2)$ and there you have it, now simplify.

28. anonymous

$cosy \frac{ dy }{ dx }=1$

29. anonymous

$\frac{ dy }{ dx }=\frac{ 1 }{ cosy }$

30. anonymous

recall

31. anonymous

Apparently my answer is wrong, terribly sorry, I'll see where I went wrong :(.

32. anonymous

$\sin^2y+\cos^2y=1$

33. anonymous

Oh yes, my sine needs to be raised to a negative 2

34. anonymous

be patient both of u i will solve it in different way and u pick up the easiest method ok, all the best

35. anonymous

oh ok

36. anonymous

$cosy=\pm \sqrt{1-\sin^2y}$

37. anonymous

thank u so much Mathsmind and its ok Sujay

38. anonymous

$\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1-\sin^2y} }$

39. anonymous

now remember our rearrangement$\sin(y)=x$ so substitute that in the above equation

40. anonymous

$\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1-x^2} }$

41. anonymous

now this is the general derivative of arcsin or sin inverse ok

42. anonymous

now if you follow the same steps for$y = \sin^{-1}(2x^3)$

43. anonymous

$\frac{ dy }{ dx } = \frac{ 6x^2 }{ \sqrt{1-4x^6} }$

44. anonymous

45. anonymous

let me prove that for you

46. anonymous

ok plzz if you can do that

47. anonymous

step One:$2x^3=siny$

48. anonymous

step two: differentiate both sides

49. anonymous

$6x^2=\frac{ dy }{ dx }cosy$

50. anonymous

step three : rearrange the differential equation

51. anonymous

$\frac{ dy }{ dx}=\frac{ 6x^2 }{ cosy }$

52. anonymous

step four: recall that $cosy=\pm \sqrt{1-\sin^2y}$

53. anonymous

step five: substitute cosy in the differential equation $\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-\sin^2y} }$

54. anonymous

wow I am liking this method better than before

55. anonymous

step six: recall$siny=2x^3s$

56. anonymous

Step seven: substitute siny in the last deferential

57. anonymous

what number is that under the power of 3?

58. anonymous

that is a mistake an s

59. anonymous

sorry i will do it again

60. anonymous

oh ok

61. anonymous

back to step six ok

62. anonymous

$siny=2x^3$

63. anonymous

ok

64. anonymous

step seven : substitute step 6 in step 5

65. anonymous

$\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-(2x^3)^2} }$

66. anonymous

finally step eight : simplify

67. anonymous

$\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-4x^6} }$

68. anonymous

and this concludes our proof good luck

69. anonymous

are you there? am sorry my computer is slow today

70. anonymous

oh wow you explained really good thanks :)

71. anonymous

so i deserve a medal then hehehe

72. anonymous

Now I have to work on 4 more of these huh lets see how am I going to do it.

73. anonymous

yup you did :)

74. anonymous

ok if you do those 4 by yourself i will give you a medal, does that sound fair?

75. anonymous

yup it does :)

76. anonymous

you can use the triangle method as well to solve this

77. anonymous

if you dont mind can I just step by step show you how to do that?

78. anonymous

sure

79. anonymous

how am i gonna give u a medal then ! hehehe

80. anonymous

thanks

81. anonymous

oh yeah sorry i forgot hehe

82. anonymous

ur welcome!

83. anonymous

u better not make one mistake

84. anonymous

ohhh hmmm I am not sure about that but I m doing it

85. anonymous

all the best

86. anonymous

there is another method using Taylor's series

87. anonymous

my professor also gave us the formula like for that equation the formula for $\sin^{-1} is 1/\sqrt{1-u ^{2}}timesdu$

88. anonymous

yes that is the method of substitution which i see it as a lazy method

89. anonymous

oh ok

90. anonymous

but if u are used to that method then go for it, as soon as u get the right answer who cares, you know!

91. anonymous

yeah

92. anonymous

i m confuse

93. anonymous

the answer that I m getting is weird

94. anonymous

are u there?

95. anonymous

yes

96. anonymous

post the question

97. anonymous

r u there?