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).

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.