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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
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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.
hmm thanks ok i ll need your help with atleast 2 of them.
bring it on !! and "WELCOME TO OPENSTUDY!!"

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Other answers:

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

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