Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Please anyone help me with this question. Find Y prime: Y=sin^1(2x^3).
 one year ago
 one year ago
Please anyone help me with this question. Find Y prime: Y=sin^1(2x^3).
 one year ago
 one year ago

This Question is Closed

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

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

goformit100Best ResponseYou've already chosen the best response.0
Yo Yo, diffrentiate on both sides...
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

JaweriaBest ResponseYou've already chosen the best response.1
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
 one year ago

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

SujayBest ResponseYou've already chosen the best response.0
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.
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

JaweriaBest ResponseYou've already chosen the best response.1
if you dont mind can I just step by step show you how to do that?
 one year ago

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

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

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

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

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

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

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

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

JaweriaBest ResponseYou've already chosen the best response.1
the answer that I m getting is weird
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.