Jaweria
I need help with 11 questions of Derivatives problems in Calculus 1. Can anyone help me with that please let me know, I will really appreciate it.
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AravindG
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we cant help with all 11 .maybe we can help in 1 or 2 with fond hope that you will understand how to do these and finish the rest yourself.
Jaweria
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hmm thanks ok i ll need your help with atleast 2 of them.
AravindG
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bring it on !! and "WELCOME TO OPENSTUDY!!"
Jaweria
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one of them is : |dw:1359270057247:dw|
Jaweria
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thank you :)
Jaweria
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and we are finding dx/dy
AravindG
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ok diff both sides with respect to y
AravindG
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use product rule
Jaweria
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ok
Jaweria
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Like this:|dw:1359270336745:dw|
Jaweria
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not sure though
Jaweria
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can anyone help me here with my questions?
nitz
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its \[x ^{2}y ^{2}=1 \] or \[xy ^{2}=1\]?
Jaweria
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its x\[its x ^{2}y ^{2}=1\]
nitz
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OK....differentiate both sides wrt to y using product rule::
\[x ^{2}(2y)+((y ^{2})2x (dx/dy))=0\]
Jaweria
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ahan ok
AravindG
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sorry i gt disconnected :O
nitz
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so tell me from above equation what would be dx/dy?
Jaweria
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oh its ok AravindG
Jaweria
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ok i will tell you
AravindG
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just solve for dx/dy after you applied product rule which will be .....
Jaweria
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I need little bit of more discription for that if its possible.
Jaweria
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everybody left me here :(
Jaweria
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no one wants to help me :(
Azteck
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\[y^2=\frac{ 1 }{ x^2 }\]
Azteck
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\[y^2=x^{-2}\]
Jaweria
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how did you get \[X ^{-2}\]
Azteck
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Do you know how to differentiate explicitly?
Azteck
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You don't know that ????
\[x^-2=\frac{1}{x^2}\]
Azteck
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\[x^-2\]
Jaweria
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oh yeah i do sorry
Azteck
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\[x^{-2}\]
Azteck
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So now, do you know how to differentiate explicitly?
Azteck
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That means you differentiate boths sides at the same time.
Jaweria
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yup
Azteck
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Okay. Give it a shot and tell me what you get. Remember, when you differentiate the y part, you also have to attach a dy/dx.
Jaweria
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ahan ok
Jaweria
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is it \[-\frac{ y }{ x}\]
Azteck
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Nope. Show me what you did as your first step.
Azteck
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I gave you this to differentiate.
\[y^2=x^{-2}\]
Jaweria
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I did like this:|dw:1359271872658:dw|
Azteck
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Then what's your next step.
Azteck
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?*
Jaweria
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then I canceled \[Y ^{2} which became Y and Then I can canceled X ^{-2} and \it became -x\]
Jaweria
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oppss sorry i did bad
nitz
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u have any doubt in my way......should i explain?
Jaweria
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lolz right now I dont have any doubt at anyone I just need help very badly :(
Azteck
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When you differentiate y^2, you get 2y?
Jaweria
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yes
Azteck
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Why did you get y?
Azteck
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and when you differentiate x^{-2}, you get -2x^{-3}
Azteck
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why did you get -x?
Jaweria
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Yes I did a mistake for -X though but I couldnt understand Y one
Azteck
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differentiate x^2
Azteck
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That's exactly the same as differentiating y^2. What are you doing?
Azteck
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\[2y\times \frac{ dy }{ dx }=-\frac{ 2 }{ x^3 }\]
Azteck
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Make dy/dx the subject.
Jaweria
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sorry one question that why there is
Jaweria
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Sorry but i have a question that how you got X^3
nitz
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\[(2x ^{2}y)=-(2xy ^{2})(dx/dy)\]
\[dx/dy=-(2x ^{2}y)/(2xy ^{2})\]
dx/dy=-x/y
Jaweria
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thanks Nitz for this :)
Jaweria
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Aztec actually the answer I am having from the professor is matching with Nitz
Azteck
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@Jaweria, do you have an answer sheet or something?
Jaweria
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yes I do
Azteck
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Ah okay. I can see that because you differentiate straight away. I moved the x^2 to the other side. Somehow that affected it.
Jaweria
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oh ok
Azteck
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nitz just differentiate straight away using the product rule.
Jaweria
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but as Nitz just explained I am little bit confused there that how he is getting 2\[2X ^{2}Y\] isnt that when you get 2X then you dont need \[2X ^{2}\]
Azteck
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Don't you know how to do the product rule?
Jaweria
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I have it on my notes and I am still trying to learn this is my first homework of this Calculus class
Jaweria
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I found it hard
Azteck
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What's the answer say in your sheet, Does it say -x/y or -y/x?
Jaweria
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it says \[-\frac{ y }{ x }\]
Azteck
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Nitz is wrong then...
Azteck
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<Look at his answer.
Azteck
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He did dx/dy. You want dy/dx
Jaweria
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yeah but I dont think so that really matters you can do either way
Jaweria
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oh yeah
Azteck
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...
Jaweria
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so its just we are flipping dy/dx instead of dx/dy
Azteck
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here's how you get -y/x
Azteck
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\[u=x^2\]
\[u'=2x\]
\[v=y^2\]
\[v'=2y\]
\[\frac{ dy }{ dx }=(u'\times v)+(u\times v')\]
Azteck
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Sorry about that.
\[0=2xy^2+2x^2y(\frac{dy}{dx})\]
Azteck
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When you differentiate y^2 you would write dy/dx next to it.
Jaweria
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yup
Azteck
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Then you make dy/dx as the subject.
Azteck
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And you're done.
Jaweria
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ahan cool thanks.
Azteck
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No worries mate.
Jaweria
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one last question plzz
Azteck
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Okay. last one. And you can try and finish the others yourself using what you got from here.
Jaweria
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alright.
Jaweria
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\[y ^{2}\tan(x)=x\]
AravindG
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to find dx/dy ?
Jaweria
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yup
AravindG
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product rule on left
AravindG
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try using draw button i will check :)
Jaweria
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hmm ok
Jaweria
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I know I ll get stuck again :( but I am trying
AravindG
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try ! try you will succeed ! :)
Jaweria
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alright :)
Jaweria
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ok first step so far: |dw:1359274476783:dw|
AravindG
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can you write thhird term again ? i cant see it clearly
AravindG
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*third
Jaweria
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ok sure
Jaweria
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|dw:1359274789970:dw|
AravindG
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i mean third term fully from y^2 sec^2 x
Jaweria
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oh ok sorry
AravindG
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and why do you have 2 equal signs ?
Jaweria
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|dw:1359274989073:dw|
Jaweria
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I put X after equal sign because thats also in the question
AravindG
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for x diff with respect to y it will be simply dx/dy isnt it?
Jaweria
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yeah thats what I pot in paranthesis next to the X
Jaweria
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put
AravindG
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you wrote dy/dx there?
Jaweria
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yes next to it
Jaweria
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|dw:1359275398794:dw|
AravindG
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what was the question ?
AravindG
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what i think is your question is \[y^2 \tan(x)=x \]
Jaweria
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\[Y ^{2}\tan (x)=x\]
AravindG
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|dw:1359275616596:dw|
AravindG
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this is what i get
Jaweria
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sorry what is equal to dy/dx?
AravindG
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nothing !!
AravindG
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we just diff with respect to dy the whole equation
Jaweria
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oh ok
AravindG
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so why do you think dy/dx will appear?
Jaweria
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so the way i did is that wrong?
AravindG
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if you wanted dx/dy I believe my method is enough ...
Jaweria
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ahan ok
AravindG
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i will be glad if someone verifies this bcoz i have not done differentiation for a while :P
Jaweria
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is Aztec here?
AravindG
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dunno , maybe he is away
AravindG
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have patience :) in the meantime try to work on other questions .BEST OF LUCK !
Jaweria
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i have an answer sheet though
Jaweria
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alright then i ll work on them :)
AravindG
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ok whats the answer you have for this question ?
Jaweria
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its \[(1-y ^{2}\sec ^{2}x)/2y \tan (x)\]
AravindG
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in btw i frgt to add tan x here
|dw:1359276176708:dw|
Jaweria
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ahan thats what I was thinking
AravindG
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i get dx/dy as reciprocal of your answer...are you sure the question asks for dx/dy ?
Jaweria
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nope its asking for dy/dx
AravindG
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then my working is right !! :)
Jaweria
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ok :)
AravindG
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solve for dx/dy from my working...dy/dx=1/dx/dy
AravindG
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understood?
Jaweria
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ok
Jaweria
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yup
Jaweria
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thanks :)
AravindG
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Ok bye I gtg
Jaweria
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ok bye
AravindG
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yw :)