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