Jaweria
  • 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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AravindG
  • AravindG
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
  • Jaweria
hmm thanks ok i ll need your help with atleast 2 of them.
AravindG
  • AravindG
bring it on !! and "WELCOME TO OPENSTUDY!!"

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Jaweria
  • Jaweria
one of them is : |dw:1359270057247:dw|
Jaweria
  • Jaweria
thank you :)
Jaweria
  • Jaweria
and we are finding dx/dy
AravindG
  • AravindG
ok diff both sides with respect to y
AravindG
  • AravindG
use product rule
Jaweria
  • Jaweria
ok
Jaweria
  • Jaweria
Like this:|dw:1359270336745:dw|
Jaweria
  • Jaweria
not sure though
Jaweria
  • Jaweria
can anyone help me here with my questions?
anonymous
  • anonymous
its \[x ^{2}y ^{2}=1 \] or \[xy ^{2}=1\]?
Jaweria
  • Jaweria
its x\[its x ^{2}y ^{2}=1\]
anonymous
  • anonymous
OK....differentiate both sides wrt to y using product rule:: \[x ^{2}(2y)+((y ^{2})2x (dx/dy))=0\]
Jaweria
  • Jaweria
ahan ok
AravindG
  • AravindG
sorry i gt disconnected :O
anonymous
  • anonymous
so tell me from above equation what would be dx/dy?
Jaweria
  • Jaweria
oh its ok AravindG
Jaweria
  • Jaweria
ok i will tell you
AravindG
  • AravindG
just solve for dx/dy after you applied product rule which will be .....
Jaweria
  • Jaweria
I need little bit of more discription for that if its possible.
Jaweria
  • Jaweria
everybody left me here :(
Jaweria
  • Jaweria
no one wants to help me :(
anonymous
  • anonymous
\[y^2=\frac{ 1 }{ x^2 }\]
anonymous
  • anonymous
\[y^2=x^{-2}\]
Jaweria
  • Jaweria
how did you get \[X ^{-2}\]
anonymous
  • anonymous
Do you know how to differentiate explicitly?
anonymous
  • anonymous
You don't know that ???? \[x^-2=\frac{1}{x^2}\]
anonymous
  • anonymous
\[x^-2\]
Jaweria
  • Jaweria
oh yeah i do sorry
anonymous
  • anonymous
\[x^{-2}\]
anonymous
  • anonymous
So now, do you know how to differentiate explicitly?
anonymous
  • anonymous
That means you differentiate boths sides at the same time.
Jaweria
  • Jaweria
yup
anonymous
  • anonymous
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
  • Jaweria
ahan ok
Jaweria
  • Jaweria
is it \[-\frac{ y }{ x}\]
anonymous
  • anonymous
Nope. Show me what you did as your first step.
anonymous
  • anonymous
I gave you this to differentiate. \[y^2=x^{-2}\]
Jaweria
  • Jaweria
I did like this:|dw:1359271872658:dw|
anonymous
  • anonymous
Then what's your next step.
anonymous
  • anonymous
?*
Jaweria
  • Jaweria
then I canceled \[Y ^{2} which became Y and Then I can canceled X ^{-2} and \it became -x\]
Jaweria
  • Jaweria
oppss sorry i did bad
anonymous
  • anonymous
u have any doubt in my way......should i explain?
Jaweria
  • Jaweria
lolz right now I dont have any doubt at anyone I just need help very badly :(
anonymous
  • anonymous
When you differentiate y^2, you get 2y?
Jaweria
  • Jaweria
yes
anonymous
  • anonymous
Why did you get y?
anonymous
  • anonymous
and when you differentiate x^{-2}, you get -2x^{-3}
anonymous
  • anonymous
why did you get -x?
Jaweria
  • Jaweria
Yes I did a mistake for -X though but I couldnt understand Y one
anonymous
  • anonymous
differentiate x^2
anonymous
  • anonymous
That's exactly the same as differentiating y^2. What are you doing?
anonymous
  • anonymous
\[2y\times \frac{ dy }{ dx }=-\frac{ 2 }{ x^3 }\]
anonymous
  • anonymous
Make dy/dx the subject.
Jaweria
  • Jaweria
sorry one question that why there is
Jaweria
  • Jaweria
Sorry but i have a question that how you got X^3
anonymous
  • anonymous
\[(2x ^{2}y)=-(2xy ^{2})(dx/dy)\] \[dx/dy=-(2x ^{2}y)/(2xy ^{2})\] dx/dy=-x/y
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=differentiate+y%5E2%3Dx%5E-2
Jaweria
  • Jaweria
thanks Nitz for this :)
Jaweria
  • Jaweria
Aztec actually the answer I am having from the professor is matching with Nitz
anonymous
  • anonymous
@Jaweria, do you have an answer sheet or something?
Jaweria
  • Jaweria
yes I do
anonymous
  • anonymous
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
  • Jaweria
oh ok
anonymous
  • anonymous
nitz just differentiate straight away using the product rule.
Jaweria
  • Jaweria
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 \[2X ^{2}\]
anonymous
  • anonymous
Don't you know how to do the product rule?
Jaweria
  • Jaweria
I have it on my notes and I am still trying to learn this is my first homework of this Calculus class
Jaweria
  • Jaweria
I found it hard
anonymous
  • anonymous
What's the answer say in your sheet, Does it say -x/y or -y/x?
Jaweria
  • Jaweria
it says \[-\frac{ y }{ x }\]
anonymous
  • anonymous
Nitz is wrong then...
anonymous
  • anonymous
anonymous
  • anonymous
He did dx/dy. You want dy/dx
Jaweria
  • Jaweria
yeah but I dont think so that really matters you can do either way
Jaweria
  • Jaweria
oh yeah
anonymous
  • anonymous
...
Jaweria
  • Jaweria
so its just we are flipping dy/dx instead of dx/dy
anonymous
  • anonymous
here's how you get -y/x
anonymous
  • anonymous
\[u=x^2\] \[u'=2x\] \[v=y^2\] \[v'=2y\] \[\frac{ dy }{ dx }=(u'\times v)+(u\times v')\]
anonymous
  • anonymous
Sorry about that. \[0=2xy^2+2x^2y(\frac{dy}{dx})\]
anonymous
  • anonymous
When you differentiate y^2 you would write dy/dx next to it.
Jaweria
  • Jaweria
yup
anonymous
  • anonymous
Then you make dy/dx as the subject.
anonymous
  • anonymous
And you're done.
Jaweria
  • Jaweria
ahan cool thanks.
anonymous
  • anonymous
No worries mate.
Jaweria
  • Jaweria
one last question plzz
anonymous
  • anonymous
Okay. last one. And you can try and finish the others yourself using what you got from here.
Jaweria
  • Jaweria
alright.
Jaweria
  • Jaweria
\[y ^{2}\tan(x)=x\]
AravindG
  • AravindG
to find dx/dy ?
Jaweria
  • Jaweria
yup
AravindG
  • AravindG
product rule on left
AravindG
  • AravindG
try using draw button i will check :)
Jaweria
  • Jaweria
hmm ok
Jaweria
  • Jaweria
I know I ll get stuck again :( but I am trying
AravindG
  • AravindG
try ! try you will succeed ! :)
Jaweria
  • Jaweria
alright :)
Jaweria
  • Jaweria
ok first step so far: |dw:1359274476783:dw|
AravindG
  • AravindG
can you write thhird term again ? i cant see it clearly
AravindG
  • AravindG
*third
Jaweria
  • Jaweria
ok sure
Jaweria
  • Jaweria
|dw:1359274789970:dw|
AravindG
  • AravindG
i mean third term fully from y^2 sec^2 x
Jaweria
  • Jaweria
oh ok sorry
AravindG
  • AravindG
and why do you have 2 equal signs ?
Jaweria
  • Jaweria
|dw:1359274989073:dw|
Jaweria
  • Jaweria
I put X after equal sign because thats also in the question
AravindG
  • AravindG
for x diff with respect to y it will be simply dx/dy isnt it?
Jaweria
  • Jaweria
yeah thats what I pot in paranthesis next to the X
Jaweria
  • Jaweria
put
AravindG
  • AravindG
you wrote dy/dx there?
Jaweria
  • Jaweria
yes next to it
Jaweria
  • Jaweria
|dw:1359275398794:dw|
AravindG
  • AravindG
what was the question ?
AravindG
  • AravindG
what i think is your question is \[y^2 \tan(x)=x \]
Jaweria
  • Jaweria
\[Y ^{2}\tan (x)=x\]
AravindG
  • AravindG
|dw:1359275616596:dw|
AravindG
  • AravindG
this is what i get
Jaweria
  • Jaweria
sorry what is equal to dy/dx?
AravindG
  • AravindG
nothing !!
AravindG
  • AravindG
we just diff with respect to dy the whole equation
Jaweria
  • Jaweria
oh ok
AravindG
  • AravindG
so why do you think dy/dx will appear?
Jaweria
  • Jaweria
so the way i did is that wrong?
AravindG
  • AravindG
if you wanted dx/dy I believe my method is enough ...
Jaweria
  • Jaweria
ahan ok
AravindG
  • AravindG
i will be glad if someone verifies this bcoz i have not done differentiation for a while :P
Jaweria
  • Jaweria
is Aztec here?
AravindG
  • AravindG
dunno , maybe he is away
AravindG
  • AravindG
have patience :) in the meantime try to work on other questions .BEST OF LUCK !
Jaweria
  • Jaweria
i have an answer sheet though
Jaweria
  • Jaweria
alright then i ll work on them :)
AravindG
  • AravindG
ok whats the answer you have for this question ?
Jaweria
  • Jaweria
its \[(1-y ^{2}\sec ^{2}x)/2y \tan (x)\]
AravindG
  • AravindG
in btw i frgt to add tan x here |dw:1359276176708:dw|
Jaweria
  • Jaweria
ahan thats what I was thinking
AravindG
  • AravindG
i get dx/dy as reciprocal of your answer...are you sure the question asks for dx/dy ?
Jaweria
  • Jaweria
nope its asking for dy/dx
AravindG
  • AravindG
then my working is right !! :)
Jaweria
  • Jaweria
ok :)
AravindG
  • AravindG
solve for dx/dy from my working...dy/dx=1/dx/dy
AravindG
  • AravindG
understood?
Jaweria
  • Jaweria
ok
Jaweria
  • Jaweria
yup
Jaweria
  • Jaweria
thanks :)
AravindG
  • AravindG
Ok bye I gtg
Jaweria
  • Jaweria
ok bye
AravindG
  • AravindG
yw :)

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