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.

- Jaweria

- jamiebookeater

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- 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

hmm thanks ok i ll need your help with atleast 2 of them.

- AravindG

bring it on !! and "WELCOME TO OPENSTUDY!!"

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## More answers

- Jaweria

one of them is : |dw:1359270057247:dw|

- Jaweria

thank you :)

- Jaweria

and we are finding dx/dy

- AravindG

ok diff both sides with respect to y

- AravindG

use product rule

- Jaweria

ok

- Jaweria

Like this:|dw:1359270336745:dw|

- Jaweria

not sure though

- Jaweria

can anyone help me here with my questions?

- anonymous

its \[x ^{2}y ^{2}=1 \] or \[xy ^{2}=1\]?

- Jaweria

its x\[its x ^{2}y ^{2}=1\]

- anonymous

OK....differentiate both sides wrt to y using product rule::
\[x ^{2}(2y)+((y ^{2})2x (dx/dy))=0\]

- Jaweria

ahan ok

- AravindG

sorry i gt disconnected :O

- anonymous

so tell me from above equation what would be dx/dy?

- Jaweria

oh its ok AravindG

- Jaweria

ok i will tell you

- AravindG

just solve for dx/dy after you applied product rule which will be .....

- Jaweria

I need little bit of more discription for that if its possible.

- Jaweria

everybody left me here :(

- Jaweria

no one wants to help me :(

- anonymous

\[y^2=\frac{ 1 }{ x^2 }\]

- anonymous

\[y^2=x^{-2}\]

- Jaweria

how did you get \[X ^{-2}\]

- anonymous

Do you know how to differentiate explicitly?

- anonymous

You don't know that ????
\[x^-2=\frac{1}{x^2}\]

- anonymous

\[x^-2\]

- Jaweria

oh yeah i do sorry

- anonymous

\[x^{-2}\]

- anonymous

So now, do you know how to differentiate explicitly?

- anonymous

That means you differentiate boths sides at the same time.

- Jaweria

yup

- 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

ahan ok

- Jaweria

is it \[-\frac{ y }{ x}\]

- anonymous

Nope. Show me what you did as your first step.

- anonymous

I gave you this to differentiate.
\[y^2=x^{-2}\]

- Jaweria

I did like this:|dw:1359271872658:dw|

- anonymous

Then what's your next step.

- anonymous

?*

- Jaweria

then I canceled \[Y ^{2} which became Y and Then I can canceled X ^{-2} and \it became -x\]

- Jaweria

oppss sorry i did bad

- anonymous

u have any doubt in my way......should i explain?

- Jaweria

lolz right now I dont have any doubt at anyone I just need help very badly :(

- anonymous

When you differentiate y^2, you get 2y?

- Jaweria

yes

- anonymous

Why did you get y?

- anonymous

and when you differentiate x^{-2}, you get -2x^{-3}

- anonymous

why did you get -x?

- Jaweria

Yes I did a mistake for -X though but I couldnt understand Y one

- anonymous

differentiate x^2

- anonymous

That's exactly the same as differentiating y^2. What are you doing?

- anonymous

\[2y\times \frac{ dy }{ dx }=-\frac{ 2 }{ x^3 }\]

- anonymous

Make dy/dx the subject.

- Jaweria

sorry one question that why there is

- Jaweria

Sorry but i have a question that how you got X^3

- anonymous

\[(2x ^{2}y)=-(2xy ^{2})(dx/dy)\]
\[dx/dy=-(2x ^{2}y)/(2xy ^{2})\]
dx/dy=-x/y

- anonymous

http://www.wolframalpha.com/input/?i=differentiate+y%5E2%3Dx%5E-2

- Jaweria

thanks Nitz for this :)

- Jaweria

Aztec actually the answer I am having from the professor is matching with Nitz

- anonymous

@Jaweria, do you have an answer sheet or something?

- Jaweria

yes I do

- 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

oh ok

- anonymous

nitz just differentiate straight away using the product rule.

- 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

Don't you know how to do the product rule?

- Jaweria

I have it on my notes and I am still trying to learn this is my first homework of this Calculus class

- Jaweria

I found it hard

- anonymous

What's the answer say in your sheet, Does it say -x/y or -y/x?

- Jaweria

it says \[-\frac{ y }{ x }\]

- anonymous

Nitz is wrong then...

- anonymous

- anonymous

He did dx/dy. You want dy/dx

- Jaweria

yeah but I dont think so that really matters you can do either way

- Jaweria

oh yeah

- anonymous

...

- Jaweria

so its just we are flipping dy/dx instead of dx/dy

- anonymous

here's how you get -y/x

- anonymous

\[u=x^2\]
\[u'=2x\]
\[v=y^2\]
\[v'=2y\]
\[\frac{ dy }{ dx }=(u'\times v)+(u\times v')\]

- anonymous

Sorry about that.
\[0=2xy^2+2x^2y(\frac{dy}{dx})\]

- anonymous

When you differentiate y^2 you would write dy/dx next to it.

- Jaweria

yup

- anonymous

Then you make dy/dx as the subject.

- anonymous

And you're done.

- Jaweria

ahan cool thanks.

- anonymous

No worries mate.

- Jaweria

one last question plzz

- anonymous

Okay. last one. And you can try and finish the others yourself using what you got from here.

- Jaweria

alright.

- Jaweria

\[y ^{2}\tan(x)=x\]

- AravindG

to find dx/dy ?

- Jaweria

yup

- AravindG

product rule on left

- AravindG

try using draw button i will check :)

- Jaweria

hmm ok

- Jaweria

I know I ll get stuck again :( but I am trying

- AravindG

try ! try you will succeed ! :)

- Jaweria

alright :)

- Jaweria

ok first step so far: |dw:1359274476783:dw|

- AravindG

can you write thhird term again ? i cant see it clearly

- AravindG

*third

- Jaweria

ok sure

- Jaweria

|dw:1359274789970:dw|

- AravindG

i mean third term fully from y^2 sec^2 x

- Jaweria

oh ok sorry

- AravindG

and why do you have 2 equal signs ?

- Jaweria

|dw:1359274989073:dw|

- Jaweria

I put X after equal sign because thats also in the question

- AravindG

for x diff with respect to y it will be simply dx/dy isnt it?

- Jaweria

yeah thats what I pot in paranthesis next to the X

- Jaweria

put

- AravindG

you wrote dy/dx there?

- Jaweria

yes next to it

- Jaweria

|dw:1359275398794:dw|

- AravindG

what was the question ?

- AravindG

what i think is your question is \[y^2 \tan(x)=x \]

- Jaweria

\[Y ^{2}\tan (x)=x\]

- AravindG

|dw:1359275616596:dw|

- AravindG

this is what i get

- Jaweria

sorry what is equal to dy/dx?

- AravindG

nothing !!

- AravindG

we just diff with respect to dy the whole equation

- Jaweria

oh ok

- AravindG

so why do you think dy/dx will appear?

- Jaweria

so the way i did is that wrong?

- AravindG

if you wanted dx/dy I believe my method is enough ...

- Jaweria

ahan ok

- AravindG

i will be glad if someone verifies this bcoz i have not done differentiation for a while :P

- Jaweria

is Aztec here?

- AravindG

dunno , maybe he is away

- AravindG

have patience :) in the meantime try to work on other questions .BEST OF LUCK !

- Jaweria

i have an answer sheet though

- Jaweria

alright then i ll work on them :)

- AravindG

ok whats the answer you have for this question ?

- Jaweria

its \[(1-y ^{2}\sec ^{2}x)/2y \tan (x)\]

- AravindG

in btw i frgt to add tan x here
|dw:1359276176708:dw|

- Jaweria

ahan thats what I was thinking

- AravindG

i get dx/dy as reciprocal of your answer...are you sure the question asks for dx/dy ?

- Jaweria

nope its asking for dy/dx

- AravindG

then my working is right !! :)

- Jaweria

ok :)

- AravindG

solve for dx/dy from my working...dy/dx=1/dx/dy

- AravindG

understood?

- Jaweria

ok

- Jaweria

yup

- Jaweria

thanks :)

- AravindG

Ok bye I gtg

- Jaweria

ok bye

- AravindG

yw :)

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