amy0799
  • amy0799
Find d2^y/dx^2 in terms of x and y.
Mathematics
katieb
  • katieb
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amy0799
  • amy0799
\[x ^{2}y ^{2}-8x=2\]
madhu.mukherjee.946
  • madhu.mukherjee.946
2xy^2+x^2(2y)dy/dx-8=0
amy0799
  • amy0799
I already have the first derivative figured out, I don't know how to find the second though

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madhu.mukherjee.946
  • madhu.mukherjee.946
just change the sides take evrything to right side except x^2(2y)dy/dx
amy0799
  • amy0799
do I just plug in x^2y^2−8x into the derivative of d2^y/dx^2?
madhu.mukherjee.946
  • madhu.mukherjee.946
nope u do like this |dw:1442671687613:dw|
madhu.mukherjee.946
  • madhu.mukherjee.946
do you understand
amy0799
  • amy0799
umm not really, is that for dy/dx?
amy0799
  • amy0799
how? I cant really read what you drew
madhu.mukherjee.946
  • madhu.mukherjee.946
okay both the drawings
amy0799
  • amy0799
can you type out ur equations please?
madhu.mukherjee.946
  • madhu.mukherjee.946
wait
madhu.mukherjee.946
  • madhu.mukherjee.946
dy/dx=8/x^2y-2xy^2/x^22y dy/dx=4/x^2y-y/x
madhu.mukherjee.946
  • madhu.mukherjee.946
now do double derivative
amy0799
  • amy0799
I got the first derivative to be \[\frac{ xy ^{2}-4 }{ x^{2}y }\]
madhu.mukherjee.946
  • madhu.mukherjee.946
nope i told the first one
madhu.mukherjee.946
  • madhu.mukherjee.946
this is not normal derivative this is implicit differentiation
amy0799
  • 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
  • madhu.mukherjee.946
yeah i'm sorry dude you are correct
madhu.mukherjee.946
  • madhu.mukherjee.946
i'm so sorry actually i've simplified it a bit
amy0799
  • amy0799
how did you simplify it?
amy0799
  • amy0799
Hello?
madhu.mukherjee.946
  • madhu.mukherjee.946
forget that what you did is absolutely correct
madhu.mukherjee.946
  • madhu.mukherjee.946
but u made a mistake with the signs it will be 4-xy^2/x^2y
amy0799
  • amy0799
oh ok. so I just find the derivative of that?
madhu.mukherjee.946
  • madhu.mukherjee.946
yes
amy0799
  • amy0799
ok. ill show u what I get in a minute
madhu.mukherjee.946
  • madhu.mukherjee.946
no problem
amy0799
  • 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
  • amy0799
r u there?
madhu.mukherjee.946
  • madhu.mukherjee.946
yes
amy0799
  • amy0799
is what I had right so far?
madhu.mukherjee.946
  • madhu.mukherjee.946
yeah
madhu.mukherjee.946
  • madhu.mukherjee.946
anything more
amy0799
  • amy0799
well I still haven't found the 2nd derivative.
amy0799
  • amy0799
@IrishBoy123 can u help?
madhu.mukherjee.946
  • madhu.mukherjee.946
dude this is the second derivative
amy0799
  • amy0799
ik but there should be more to it
madhu.mukherjee.946
  • madhu.mukherjee.946
see if you are able to simplify it more but this is f9
amy0799
  • amy0799
f9?
madhu.mukherjee.946
  • madhu.mukherjee.946
yeah i mean fine
madhu.mukherjee.946
  • madhu.mukherjee.946
this is correct
amy0799
  • amy0799
don't I need to plug in 4-xy^2/x^2y into dy/dx?
madhu.mukherjee.946
  • madhu.mukherjee.946
yes
amy0799
  • amy0799
I got \[\frac{ 2x^2 y^5-8xy^3-16}{ y}\] but that's the wrong answer when I summited it
IrishBoy123
  • IrishBoy123
your first deriv looks wrong way round for starters
amy0799
  • amy0799
what is it suppose to be?
IrishBoy123
  • IrishBoy123
should be \[\color{red} - \frac{ xy ^{2}-4 }{ x^{2}y }\]
madhu.mukherjee.946
  • madhu.mukherjee.946
nope the first derivative done by amy is f9
madhu.mukherjee.946
  • madhu.mukherjee.946
i've checked that in wolfram alpha
IrishBoy123
  • 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
  • madhu.mukherjee.946
????????
amy0799
  • amy0799
@IrishBoy123 how would I find the answer?
madhu.mukherjee.946
  • madhu.mukherjee.946
your second derivative is wrong
madhu.mukherjee.946
  • madhu.mukherjee.946
y^2(y-3xy'(x))/x^2
IrishBoy123
  • 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
  • amy0799
isn't that gonna be a big number when I plug in y'?
IrishBoy123
  • 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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