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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0When you say y prime, do you mean the derivative?

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1Please someone help me I m stuck and its due tomorrow

goformit100
 3 years ago
Best ResponseYou've already chosen the best response.0Yo Yo, diffrentiate on both sides...

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1These are kind of new problems to me if you dont mind can you teach me and help me solving them?

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1can we go step by step?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ya sure, I'll set it up.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you want to go by steps sure will do

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0at the end i will show u how u can work such equation in ur head, without paper and pen

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks to both of you Sujay and Mathsmind :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Original Equation:\[y=\sin^21(2x3)?\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0with an x after the sin

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I want to make sure my original is correct

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[y=\sin^{1}(2x^3)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0please confirm is this what u r asking for?

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1ok let me write down the equation here. Its \[y=\sin^{1} (2x ^{3})\]

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1yeah I m here sorry I was just writing that equation down

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1yeah you are right Mathmind

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok there are 2 main ways to solve this problem at ur level

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1is 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
 3 years ago
Best ResponseYou've already chosen the best response.0differentiate both sides

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Original:\[\sin^{1} (2x^3)\] 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[cosy \frac{ dy }{ dx }=1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx }=\frac{ 1 }{ cosy }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Apparently my answer is wrong, terribly sorry, I'll see where I went wrong :(.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sin^2y+\cos^2y=1\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh yes, my sine needs to be raised to a negative 2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0be patient both of u i will solve it in different way and u pick up the easiest method ok, all the best

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[cosy=\pm \sqrt{1\sin^2y}\]

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1thank u so much Mathsmind and its ok Sujay

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1\sin^2y} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now remember our rearrangement\[\sin(y)=x\] so substitute that in the above equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1x^2} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now this is the general derivative of arcsin or sin inverse ok

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now if you follow the same steps for\[y = \sin^{1}(2x^3)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx } = \frac{ 6x^2 }{ \sqrt{14x^6} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and that's is your final answer achieved using the chain rule

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0let me prove that for you

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1ok plzz if you can do that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0step One:\[2x^3=siny\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0step two: differentiate both sides

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[6x^2=\frac{ dy }{ dx }cosy\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0step three : rearrange the differential equation

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx}=\frac{ 6x^2 }{ cosy }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0step four: recall that \[cosy=\pm \sqrt{1\sin^2y}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0step five: substitute cosy in the differential equation \[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1\sin^2y} }\]

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1wow I am liking this method better than before

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0step six: recall\[siny=2x^3s\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Step seven: substitute siny in the last deferential

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1what number is that under the power of 3?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that is a mistake an s

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry i will do it again

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0step seven : substitute step 6 in step 5

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{1(2x^3)^2} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0finally step eight : simplify

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{14x^6} }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and this concludes our proof good luck

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0are you there? am sorry my computer is slow today

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1oh wow you explained really good thanks :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i deserve a medal then hehehe

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1Now I have to work on 4 more of these huh lets see how am I going to do it.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok if you do those 4 by yourself i will give you a medal, does that sound fair?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you can use the triangle method as well to solve this

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1if you dont mind can I just step by step show you how to do that?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how am i gonna give u a medal then ! hehehe

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1oh yeah sorry i forgot hehe

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u better not make one mistake

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1ohhh hmmm I am not sure about that but I m doing it

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there is another method using Taylor's series

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1my professor also gave us the formula like for that equation the formula for \[\sin^{1} is 1/\sqrt{1u ^{2}}timesdu\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes that is the method of substitution which i see it as a lazy method

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but if u are used to that method then go for it, as soon as u get the right answer who cares, you know!

jaweria
 3 years ago
Best ResponseYou've already chosen the best response.1the answer that I m getting is weird
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