- amy0799

Find d2^y/dx^2 in terms of x and y.

- katieb

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- amy0799

\[x ^{2}y ^{2}-8x=2\]

- madhu.mukherjee.946

2xy^2+x^2(2y)dy/dx-8=0

- amy0799

I already have the first derivative figured out, I don't know how to find the second though

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## More answers

- madhu.mukherjee.946

just change the sides take evrything to right side except x^2(2y)dy/dx

- amy0799

do I just plug in x^2y^2−8x into the derivative of d2^y/dx^2?

- madhu.mukherjee.946

nope u do like this |dw:1442671687613:dw|

- madhu.mukherjee.946

do you understand

- amy0799

umm not really, is that for dy/dx?

- amy0799

how? I cant really read what you drew

- madhu.mukherjee.946

okay both the drawings

- amy0799

can you type out ur equations please?

- madhu.mukherjee.946

wait

- madhu.mukherjee.946

dy/dx=8/x^2y-2xy^2/x^22y
dy/dx=4/x^2y-y/x

- madhu.mukherjee.946

now do double derivative

- amy0799

I got the first derivative to be \[\frac{ xy ^{2}-4 }{ x^{2}y }\]

- madhu.mukherjee.946

nope i told the first one

- madhu.mukherjee.946

this is not normal derivative this is implicit differentiation

- amy0799

I know but when I did the implicit differentiation for x^2y^2−8x=2 I got that equation and it's different than yours

- madhu.mukherjee.946

yeah i'm sorry dude you are correct

- madhu.mukherjee.946

i'm so sorry actually i've simplified it a bit

- amy0799

how did you simplify it?

- amy0799

Hello?

- madhu.mukherjee.946

forget that what you did is absolutely correct

- madhu.mukherjee.946

but u made a mistake with the signs it will be 4-xy^2/x^2y

- amy0799

oh ok. so I just find the derivative of that?

- madhu.mukherjee.946

yes

- amy0799

ok. ill show u what I get in a minute

- madhu.mukherjee.946

no problem

- amy0799

I got \[\frac{ x ^{2}y(-2xyy'-y ^{2})-(4-xy ^{2})(x ^{2}y'+2xy) }{ y ^{2}x ^{4} }\]
correct so far?

- amy0799

r u there?

- madhu.mukherjee.946

yes

- amy0799

is what I had right so far?

- madhu.mukherjee.946

yeah

- madhu.mukherjee.946

anything more

- amy0799

well I still haven't found the 2nd derivative.

- amy0799

@IrishBoy123 can u help?

- madhu.mukherjee.946

dude this is the second derivative

- amy0799

ik but there should be more to it

- madhu.mukherjee.946

see if you are able to simplify it more but this is f9

- amy0799

f9?

- madhu.mukherjee.946

yeah i mean fine

- madhu.mukherjee.946

this is correct

- amy0799

don't I need to plug in 4-xy^2/x^2y into dy/dx?

- madhu.mukherjee.946

yes

- amy0799

I got \[\frac{ 2x^2 y^5-8xy^3-16}{ y}\]
but that's the wrong answer when I summited it

- IrishBoy123

your first deriv looks wrong way round for starters

- amy0799

what is it suppose to be?

- IrishBoy123

should be \[\color{red} - \frac{ xy ^{2}-4 }{ x^{2}y }\]

- madhu.mukherjee.946

nope the first derivative done by amy is f9

- madhu.mukherjee.946

i've checked that in wolfram alpha

- IrishBoy123

you can then do it implicitly starting
\[y' .x^2 .y = 4 - xy^2 \]
and applying the triple product, see here
|dw:1442677649641:dw|
but, whatever way you do it, it's horrible.

- madhu.mukherjee.946

????????

- amy0799

@IrishBoy123 how would I find the answer?

- madhu.mukherjee.946

your second derivative is wrong

- madhu.mukherjee.946

y^2(y-3xy'(x))/x^2

- IrishBoy123

first deriv
\[x ^{2}y ^{2}-8x=2\]
\[2x y ^{2} + x^22yy'-8 = 0\]
\[y' = \frac{8-2xy^2}{2x^2y} = ....\]ie wrong
for second deriv you **might** go:
\[ (x^2)(y)(y')=4 -x y ^{2} \]
\[ (x^2)'(y)(y') +(x^2)(y)'(y')+(x^2)(y)(y')' = -y^2-2xyy'\]
at some point you plug in for y'

- amy0799

isn't that gonna be a big number when I plug in y'?

- IrishBoy123

it's going to be horrible whatever way you do it. i have a feeling this could be a more organised way to do it, but if you are most of the way through, carry on as before
however, do check that first derivative again or you will never get there

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