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

- Jaweria

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

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- schrodinger

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- anonymous

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

- Jaweria

yup

- Jaweria

Please someone help me I m stuck and its due tomorrow

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## More answers

- goformit100

Yo Yo, diffrentiate on both sides...

- Jaweria

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

- anonymous

the answer is:

- Jaweria

can we go step by step?

- anonymous

Ya sure, I'll set it up.

- anonymous

you want to go by steps sure will do

- anonymous

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

- Jaweria

Thanks to both of you Sujay and Mathsmind :)

- anonymous

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

- anonymous

with an x after the sin

- anonymous

I want to make sure my original is correct

- anonymous

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

- anonymous

please confirm is this what u r asking for?

- anonymous

are u there?

- Jaweria

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

- anonymous

i did write it

- Jaweria

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

- Jaweria

yeah you are right Mathmind

- anonymous

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

- 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

- anonymous

\[y = \arcsin(x)\]

- anonymous

\[\sin(y)=x\]

- anonymous

differentiate both sides

- 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.

- anonymous

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

- anonymous

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

- anonymous

recall

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- Jaweria

oh ok

- anonymous

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

- Jaweria

thank u so much Mathsmind and its ok Sujay

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

let me prove that for you

- Jaweria

ok plzz if you can do that

- anonymous

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

- anonymous

step two: differentiate both sides

- anonymous

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

- anonymous

step three : rearrange the differential equation

- anonymous

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

- anonymous

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

- anonymous

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

- Jaweria

wow I am liking this method better than before

- anonymous

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

- anonymous

Step seven: substitute siny in the last deferential

- Jaweria

what number is that under the power of 3?

- anonymous

that is a mistake an s

- anonymous

sorry i will do it again

- Jaweria

oh ok

- anonymous

back to step six ok

- anonymous

\[siny=2x^3\]

- Jaweria

ok

- anonymous

step seven : substitute step 6 in step 5

- anonymous

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

- anonymous

finally step eight : simplify

- anonymous

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

- anonymous

and this concludes our proof good luck

- anonymous

are you there? am sorry my computer is slow today

- Jaweria

oh wow you explained really good thanks :)

- anonymous

so i deserve a medal then hehehe

- Jaweria

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

- Jaweria

yup you did :)

- anonymous

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

- Jaweria

yup it does :)

- anonymous

you can use the triangle method as well to solve this

- Jaweria

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

- anonymous

sure

- anonymous

how am i gonna give u a medal then ! hehehe

- Jaweria

thanks

- Jaweria

oh yeah sorry i forgot hehe

- anonymous

ur welcome!

- anonymous

u better not make one mistake

- Jaweria

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

- anonymous

all the best

- anonymous

there is another method using Taylor's series

- Jaweria

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

- anonymous

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

- Jaweria

oh ok

- 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!

- Jaweria

yeah

- Jaweria

i m confuse

- Jaweria

the answer that I m getting is weird

- Jaweria

are u there?

- anonymous

yes

- anonymous

post the question

- anonymous

r u there?

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