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Sujay
 one year ago
Best ResponseYou've already chosen the best response.0When you say y prime, do you mean the derivative?

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1Please someone help me I m stuck and its due tomorrow

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

Jaweria
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1can we go step by step?

Sujay
 one year ago
Best ResponseYou've already chosen the best response.0Ya sure, I'll set it up.

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1you want to go by steps sure will do

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1at the end i will show u how u can work such equation in ur head, without paper and pen

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1Thanks to both of you Sujay and Mathsmind :)

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

Sujay
 one year ago
Best ResponseYou've already chosen the best response.0I want to make sure my original is correct

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1please confirm is this what u r asking for?

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

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1yeah I m here sorry I was just writing that equation down

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1yeah you are right Mathmind

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1ok there are 2 main ways to solve this problem at ur level

Jaweria
 one year 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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1differentiate both sides

Sujay
 one year 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.

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[cosy \frac{ dy }{ dx }=1\]

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dx }=\frac{ 1 }{ cosy }\]

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[\sin^2y+\cos^2y=1\]

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1be patient both of u i will solve it in different way and u pick up the easiest method ok, all the best

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[cosy=\pm \sqrt{1\sin^2y}\]

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1thank u so much Mathsmind and its ok Sujay

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1\sin^2y} }\]

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1now remember our rearrangement\[\sin(y)=x\] so substitute that in the above equation

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dx } = \frac{ 1 }{ \sqrt{1x^2} }\]

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1now this is the general derivative of arcsin or sin inverse ok

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dx } = \frac{ 6x^2 }{ \sqrt{14x^6} }\]

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1and that's is your final answer achieved using the chain rule

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1let me prove that for you

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1ok plzz if you can do that

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1step One:\[2x^3=siny\]

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1step two: differentiate both sides

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[6x^2=\frac{ dy }{ dx }cosy\]

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1step three : rearrange the differential equation

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dx}=\frac{ 6x^2 }{ cosy }\]

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

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

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1wow I am liking this method better than before

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1step six: recall\[siny=2x^3s\]

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1Step seven: substitute siny in the last deferential

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1that is a mistake an s

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1sorry i will do it again

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1step seven : substitute step 6 in step 5

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1finally step eight : simplify

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ dy }{ dx}=\frac{ 6x^2 }{ \sqrt{14x^6} }\]

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1and this concludes our proof good luck

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1are you there? am sorry my computer is slow today

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1oh wow you explained really good thanks :)

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1so i deserve a medal then hehehe

Jaweria
 one year 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.

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1ok if you do those 4 by yourself i will give you a medal, does that sound fair?

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1you can use the triangle method as well to solve this

Jaweria
 one year 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?

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1how am i gonna give u a medal then ! hehehe

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1oh yeah sorry i forgot hehe

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1u better not make one mistake

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1there is another method using Taylor's series

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

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1yes that is the method of substitution which i see it as a lazy method

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.1but if u are used to that method then go for it, as soon as u get the right answer who cares, you know!

Jaweria
 one year ago
Best ResponseYou've already chosen the best response.1the answer that I m getting is weird
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