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

Calculus1
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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

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Other answers:

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
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
\[y = \arcsin(x)\]
\[\sin(y)=x\]
differentiate both sides
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.
\[cosy \frac{ dy }{ dx }=1\]
\[\frac{ dy }{ dx }=\frac{ 1 }{ cosy }\]
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
be patient both of u i will solve it in different way and u pick up the easiest method ok, all the best
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 \[cosy=\pm \sqrt{1-\sin^2y}\]
step five: substitute cosy in the differential equation \[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-\sin^2y} }\]
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
my professor also gave us the formula like for that equation the formula for \[\sin^{-1} is 1/\sqrt{1-u ^{2}}timesdu\]
yes that is the method of substitution which i see it as a lazy method
oh ok
but if u are used to that method then go for it, as soon as u get the right answer who cares, you know!
yeah
i m confuse
the answer that I m getting is weird
are u there?
yes
post the question
r u there?

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