Jaweria
  • Jaweria
Please anyone help me with this question. Find Y prime: Y=sin^-1(2x^3).
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
When you say y prime, do you mean the derivative?
Jaweria
  • Jaweria
yup
Jaweria
  • Jaweria
Please someone help me I m stuck and its due tomorrow

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More answers

goformit100
  • goformit100
Yo Yo, diffrentiate on both sides...
Jaweria
  • Jaweria
These are kind of new problems to me if you dont mind can you teach me and help me solving them?
anonymous
  • anonymous
the answer is:
Jaweria
  • Jaweria
can we go step by step?
anonymous
  • anonymous
Ya sure, I'll set it up.
anonymous
  • anonymous
you want to go by steps sure will do
anonymous
  • anonymous
at the end i will show u how u can work such equation in ur head, without paper and pen
Jaweria
  • Jaweria
Thanks to both of you Sujay and Mathsmind :)
anonymous
  • anonymous
Original Equation:\[y=\sin^2-1(2x-3)?\]
anonymous
  • anonymous
with an x after the sin
anonymous
  • anonymous
I want to make sure my original is correct
anonymous
  • anonymous
\[y=\sin^{-1}(2x^3)\]
anonymous
  • anonymous
please confirm is this what u r asking for?
anonymous
  • anonymous
are u there?
Jaweria
  • Jaweria
ok let me write down the equation here. Its \[y=\sin^{-1} (2x ^{3})\]
anonymous
  • anonymous
i did write it
Jaweria
  • Jaweria
yeah I m here sorry I was just writing that equation down
Jaweria
  • Jaweria
yeah you are right Mathmind
anonymous
  • anonymous
ok there are 2 main ways to solve this problem at ur level
Jaweria
  • 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
  • anonymous
\[y = \arcsin(x)\]
anonymous
  • anonymous
\[\sin(y)=x\]
anonymous
  • anonymous
differentiate both sides
anonymous
  • 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
  • anonymous
\[cosy \frac{ dy }{ dx }=1\]
anonymous
  • anonymous
\[\frac{ dy }{ dx }=\frac{ 1 }{ cosy }\]
anonymous
  • anonymous
recall
anonymous
  • anonymous
Apparently my answer is wrong, terribly sorry, I'll see where I went wrong :(.
anonymous
  • anonymous
\[\sin^2y+\cos^2y=1\]
anonymous
  • anonymous
Oh yes, my sine needs to be raised to a negative 2
anonymous
  • 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
  • Jaweria
oh ok
anonymous
  • anonymous
\[cosy=\pm \sqrt{1-\sin^2y}\]
Jaweria
  • Jaweria
thank u so much Mathsmind and its ok Sujay
anonymous
  • anonymous
\[\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1-\sin^2y} }\]
anonymous
  • anonymous
now remember our rearrangement\[\sin(y)=x\] so substitute that in the above equation
anonymous
  • anonymous
\[\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1-x^2} }\]
anonymous
  • anonymous
now this is the general derivative of arcsin or sin inverse ok
anonymous
  • anonymous
now if you follow the same steps for\[y = \sin^{-1}(2x^3)\]
anonymous
  • anonymous
\[\frac{ dy }{ dx } = \frac{ 6x^2 }{ \sqrt{1-4x^6} }\]
anonymous
  • anonymous
and that's is your final answer achieved using the chain rule
anonymous
  • anonymous
let me prove that for you
Jaweria
  • Jaweria
ok plzz if you can do that
anonymous
  • anonymous
step One:\[2x^3=siny\]
anonymous
  • anonymous
step two: differentiate both sides
anonymous
  • anonymous
\[6x^2=\frac{ dy }{ dx }cosy\]
anonymous
  • anonymous
step three : rearrange the differential equation
anonymous
  • anonymous
\[\frac{ dy }{ dx}=\frac{ 6x^2 }{ cosy }\]
anonymous
  • anonymous
step four: recall that \[cosy=\pm \sqrt{1-\sin^2y}\]
anonymous
  • anonymous
step five: substitute cosy in the differential equation \[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-\sin^2y} }\]
Jaweria
  • Jaweria
wow I am liking this method better than before
anonymous
  • anonymous
step six: recall\[siny=2x^3s\]
anonymous
  • anonymous
Step seven: substitute siny in the last deferential
Jaweria
  • Jaweria
what number is that under the power of 3?
anonymous
  • anonymous
that is a mistake an s
anonymous
  • anonymous
sorry i will do it again
Jaweria
  • Jaweria
oh ok
anonymous
  • anonymous
back to step six ok
anonymous
  • anonymous
\[siny=2x^3\]
Jaweria
  • Jaweria
ok
anonymous
  • anonymous
step seven : substitute step 6 in step 5
anonymous
  • anonymous
\[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-(2x^3)^2} }\]
anonymous
  • anonymous
finally step eight : simplify
anonymous
  • anonymous
\[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1-4x^6} }\]
anonymous
  • anonymous
and this concludes our proof good luck
anonymous
  • anonymous
are you there? am sorry my computer is slow today
Jaweria
  • Jaweria
oh wow you explained really good thanks :)
anonymous
  • anonymous
so i deserve a medal then hehehe
Jaweria
  • Jaweria
Now I have to work on 4 more of these huh lets see how am I going to do it.
Jaweria
  • Jaweria
yup you did :)
anonymous
  • anonymous
ok if you do those 4 by yourself i will give you a medal, does that sound fair?
Jaweria
  • Jaweria
yup it does :)
anonymous
  • anonymous
you can use the triangle method as well to solve this
Jaweria
  • Jaweria
if you dont mind can I just step by step show you how to do that?
anonymous
  • anonymous
sure
anonymous
  • anonymous
how am i gonna give u a medal then ! hehehe
Jaweria
  • Jaweria
thanks
Jaweria
  • Jaweria
oh yeah sorry i forgot hehe
anonymous
  • anonymous
ur welcome!
anonymous
  • anonymous
u better not make one mistake
Jaweria
  • Jaweria
ohhh hmmm I am not sure about that but I m doing it
anonymous
  • anonymous
all the best
anonymous
  • anonymous
there is another method using Taylor's series
Jaweria
  • 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
  • anonymous
yes that is the method of substitution which i see it as a lazy method
Jaweria
  • Jaweria
oh ok
anonymous
  • 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
  • Jaweria
yeah
Jaweria
  • Jaweria
i m confuse
Jaweria
  • Jaweria
the answer that I m getting is weird
Jaweria
  • Jaweria
are u there?
anonymous
  • anonymous
yes
anonymous
  • anonymous
post the question
anonymous
  • anonymous
r u there?

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