Question

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

Answer 1

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

Answer 2

yup

Answer 3

Please someone help me I m stuck and its due tomorrow

Answer 4

Yo Yo, diffrentiate on both sides...

Answer 5

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

Answer 6

the answer is:

Answer 7

can we go step by step?

Answer 8

Ya sure, I'll set it up.

Answer 9

you want to go by steps sure will do

Answer 10

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

Answer 11

Thanks to both of you Sujay and Mathsmind :)

Answer 12

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

Answer 13

with an x after the sin

Answer 14

I want to make sure my original is correct

Answer 15

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

Answer 16

please confirm is this what u r asking for?

Answer 17

are u there?

Answer 18

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

Answer 19

i did write it

Answer 20

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

Answer 21

yeah you are right Mathmind

Answer 22

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

Answer 23

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

Answer 24

\[y = \arcsin(x)\]

Answer 25

\[\sin(y)=x\]

Answer 26

differentiate both sides

Answer 27

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.

Answer 28

\[cosy \frac{ dy }{ dx }=1\]

Answer 29

\[\frac{ dy }{ dx }=\frac{ 1 }{ cosy }\]

Answer 30

recall

Answer 31

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

Answer 32

\[\sin^2y+\cos^2y=1\]

Answer 33

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

Answer 34

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

Answer 35

oh ok

Answer 36

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

Answer 37

thank u so much Mathsmind and its ok Sujay

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

and that's is your final answer achieved using the chain rule

Answer 45

let me prove that for you

Answer 46

ok plzz if you can do that

Answer 47

step One:\[2x^3=siny\]

Answer 48

step two: differentiate both sides

Answer 49

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

Answer 50

step three : rearrange the differential equation

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

wow I am liking this method better than before

Answer 55

step six: recall\[siny=2x^3s\]

Answer 56

Step seven: substitute siny in the last deferential

Answer 57

what number is that under the power of 3?

Answer 58

that is a mistake an s

Answer 59

sorry i will do it again

Answer 60

oh ok

Answer 61

back to step six ok

Answer 62

\[siny=2x^3\]

Answer 63

ok

Answer 64

step seven : substitute step 6 in step 5

Answer 65

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

Answer 66

finally step eight : simplify

Answer 67

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

Answer 68

and this concludes our proof good luck

Answer 69

are you there? am sorry my computer is slow today

Answer 70

oh wow you explained really good thanks :)

Answer 71

so i deserve a medal then hehehe

Answer 72

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

Answer 73

yup you did :)

Answer 74

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

Answer 75

yup it does :)

Answer 76

you can use the triangle method as well to solve this

Answer 77

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

Answer 78

sure

Answer 79

how am i gonna give u a medal then ! hehehe

Answer 80

thanks

Answer 81

oh yeah sorry i forgot hehe

Answer 82

ur welcome!

Answer 83

u better not make one mistake

Answer 84

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

Answer 85

all the best

Answer 86

there is another method using Taylor's series

Answer 87

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

Answer 88

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

Answer 89

oh ok

Answer 90

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

Answer 91

yeah

Answer 92

i m confuse

Answer 93

the answer that I m getting is weird

Answer 94

are u there?

Answer 95

yes

Answer 96

post the question

Answer 97

r u there?