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Double Implicit differantion, So I got this: http://screencast.com/t/HZNKpdBC Now I have found that dy/dx = 2x/3y How can I find d^2y/dx^2 ?

Mathematics
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If \[\frac{dy}{dx}=\frac{2x}{3y}\]Then you just need to take the partial derivative of that to get the 2nd derivative.
partial?

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I solved this now I am trying to solve this: http://screencast.com/t/tlp6QrmBh7 Here is the solution: But my result is: (2y^3x+2x^4)/y^5
Solution: http://screencast.com/t/ZKI6Di9vkR8j ***
Take your solution and factor an x out to get theirs.
still no
Okay, so this is what you have so far, right?\[x^{3}+y^{3}=1\rightarrow 3x^{2}+3y^{2}\frac{dy}{dx}=0\rightarrow \frac{dy}{dx}=-\frac{3x^{2}}{3y^{2}}\rightarrow \frac{dy}{dx}=-\frac{x^{2}}{y^{2}}\]
yes
Then you take the derivative of that using the quotient rule of \[[\frac{f(x)}{g(x)}]'=\frac{g(x)f(x)'-f(x)g(x)'}{[g(x)]^{2}}\] which is \[\frac{y^2(2x)-x^2(2y)\frac{dy}{dx}}{[y^{2}]^{2}}\]
yay
It's ok guys I think am gonna surpass this, I guess my answer is similar enought if not 100% correct , am gonna do some derivatives and head to sleep
Seems I get the same answer you did, not sure where they are getting that extra x.
\[2x^{2}-3y ^{2}=4\] \[2*2x-3*2y \frac{ dy }{dx }=0\] \[2x-3y \frac{ dy }{dx }=0,\frac{ dy }{dx }=\frac{ 2x }{ 3y }\] \[\frac{ d ^{2}y }{dx ^{2} }=\frac{ 2 }{3 }\frac{y*1-x \frac{ dy }{dx } }{ y ^{2} }\] \[\frac{ d ^{2}y }{dx ^{2} }=\frac{ 2 }{3 }\frac{y-x \frac{ 2x }{3y } }{y ^{2} }\] \[\frac{ d ^{2}y}{dx ^{2} }=\frac{ 2 }{ 3 }\frac{ \frac{ 3y ^{2}-2x ^{2} }{ 3y } }{y ^{2} }\] \[\frac{ d ^{2}y }{ dx ^{2} }=-\frac{ 2 }{9 }\frac{ 2x ^{2}-3y ^{2} }{ y ^{3} }\] \[\frac{ d ^{2}y }{ dx ^{2} }=-\frac{ 2 }{9 }\frac{ 4 }{y ^{3} }\] \[\frac{ d ^{2}y }{dx ^{2} }=-\frac{ 8 }{9y ^{3} }\]

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