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

yup

Please someone help me I m stuck and its due tomorrow

Yo Yo, diffrentiate on both sides...

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

the answer is:

can we go step by step?

Ya sure, I'll set it up.

you want to go by steps sure will do

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

Thanks to both of you Sujay and Mathsmind :)

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

with an x after the sin

I want to make sure my original is correct

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

please confirm is this what u r asking for?

are u there?

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

i did write it

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

yeah you are right Mathmind

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

\[y = \arcsin(x)\]

\[\sin(y)=x\]

differentiate both sides

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

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

recall

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

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

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

oh ok

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

thank u so much Mathsmind and its ok Sujay

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

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

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

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

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

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

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

let me prove that for you

ok plzz if you can do that

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

step two: differentiate both sides

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

step three : rearrange the differential equation

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

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

wow I am liking this method better than before

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

Step seven: substitute siny in the last deferential

what number is that under the power of 3?

that is a mistake an s

sorry i will do it again

oh ok

back to step six ok

\[siny=2x^3\]

ok

step seven : substitute step 6 in step 5

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

finally step eight : simplify

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

and this concludes our proof good luck

are you there? am sorry my computer is slow today

oh wow you explained really good thanks :)

so i deserve a medal then hehehe

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

yup you did :)

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

yup it does :)

you can use the triangle method as well to solve this

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

sure

how am i gonna give u a medal then ! hehehe

thanks

oh yeah sorry i forgot hehe

ur welcome!

u better not make one mistake

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

all the best

there is another method using Taylor's series

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

oh ok

yeah

i m confuse

the answer that I m getting is weird

are u there?

yes

post the question

r u there?