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

1. Sujay

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

2. Jaweria

yup

3. Jaweria

Please someone help me I m stuck and its due tomorrow

4. goformit100

Yo Yo, diffrentiate on both sides...

5. Jaweria

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

6. mathsmind

7. Jaweria

can we go step by step?

8. Sujay

Ya sure, I'll set it up.

9. mathsmind

you want to go by steps sure will do

10. mathsmind

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

11. Jaweria

Thanks to both of you Sujay and Mathsmind :)

12. Sujay

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

13. Sujay

with an x after the sin

14. Sujay

I want to make sure my original is correct

15. mathsmind

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

16. mathsmind

17. mathsmind

are u there?

18. Jaweria

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

19. mathsmind

i did write it

20. Jaweria

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

21. Jaweria

yeah you are right Mathmind

22. mathsmind

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

23. Jaweria

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. mathsmind

$y = \arcsin(x)$

25. mathsmind

$\sin(y)=x$

26. mathsmind

differentiate both sides

27. Sujay

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. mathsmind

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

29. mathsmind

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

30. mathsmind

recall

31. Sujay

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

32. mathsmind

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

33. Sujay

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

34. mathsmind

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

35. Jaweria

oh ok

36. mathsmind

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

37. Jaweria

thank u so much Mathsmind and its ok Sujay

38. mathsmind

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

39. mathsmind

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

40. mathsmind

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

41. mathsmind

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

42. mathsmind

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

43. mathsmind

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

44. mathsmind

45. mathsmind

let me prove that for you

46. Jaweria

ok plzz if you can do that

47. mathsmind

step One:$2x^3=siny$

48. mathsmind

step two: differentiate both sides

49. mathsmind

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

50. mathsmind

step three : rearrange the differential equation

51. mathsmind

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

52. mathsmind

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

53. mathsmind

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

54. Jaweria

wow I am liking this method better than before

55. mathsmind

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

56. mathsmind

Step seven: substitute siny in the last deferential

57. Jaweria

what number is that under the power of 3?

58. mathsmind

that is a mistake an s

59. mathsmind

sorry i will do it again

60. Jaweria

oh ok

61. mathsmind

back to step six ok

62. mathsmind

$siny=2x^3$

63. Jaweria

ok

64. mathsmind

step seven : substitute step 6 in step 5

65. mathsmind

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

66. mathsmind

finally step eight : simplify

67. mathsmind

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

68. mathsmind

and this concludes our proof good luck

69. mathsmind

are you there? am sorry my computer is slow today

70. Jaweria

oh wow you explained really good thanks :)

71. mathsmind

so i deserve a medal then hehehe

72. Jaweria

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

73. Jaweria

yup you did :)

74. mathsmind

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

75. Jaweria

yup it does :)

76. mathsmind

you can use the triangle method as well to solve this

77. Jaweria

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

78. mathsmind

sure

79. mathsmind

how am i gonna give u a medal then ! hehehe

80. Jaweria

thanks

81. Jaweria

oh yeah sorry i forgot hehe

82. mathsmind

ur welcome!

83. mathsmind

u better not make one mistake

84. Jaweria

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

85. mathsmind

all the best

86. mathsmind

there is another method using Taylor's series

87. Jaweria

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

88. mathsmind

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

89. Jaweria

oh ok

90. mathsmind

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. Jaweria

yeah

92. Jaweria

i m confuse

93. Jaweria

the answer that I m getting is weird

94. Jaweria

are u there?

95. mathsmind

yes

96. mathsmind

post the question

97. mathsmind

r u there?