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Christos

  • 2 years ago

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 ?

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  1. Christos
    • 2 years ago
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    @hero @jim @Luigi0210 @EsaaLoca @eSpeX @Emily778 @timo86m @zzr0ck3r

  2. eSpeX
    • 2 years ago
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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.

  3. zzr0ck3r
    • 2 years ago
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    partial?

  4. Christos
    • 2 years ago
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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

  5. Christos
    • 2 years ago
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    Solution: http://screencast.com/t/ZKI6Di9vkR8j ***

  6. eSpeX
    • 2 years ago
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    Take your solution and factor an x out to get theirs.

  7. Christos
    • 2 years ago
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    still no

  8. eSpeX
    • 2 years ago
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    Okay, so this is what you have so far, right?\[x^{3}+y^{3}=1\rightarrow 3x^{2}+3y^{2}\frac{dy}{dx}=0\rightarrow \frac{dy}{dx}=-\frac{3x^{2}}{3y^{2}}\rightarrow \frac{dy}{dx}=-\frac{x^{2}}{y^{2}}\]

  9. Christos
    • 2 years ago
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    yes

  10. eSpeX
    • 2 years ago
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    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 \[\frac{y^2(2x)-x^2(2y)\frac{dy}{dx}}{[y^{2}]^{2}}\]

  11. Christos
    • 2 years ago
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    yay

  12. Christos
    • 2 years ago
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    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

  13. eSpeX
    • 2 years ago
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    Seems I get the same answer you did, not sure where they are getting that extra x.

  14. surjithayer
    • 2 years ago
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    \[2x^{2}-3y ^{2}=4\] \[2*2x-3*2y \frac{ dy }{dx }=0\] \[2x-3y \frac{ dy }{dx }=0,\frac{ dy }{dx }=\frac{ 2x }{ 3y }\] \[\frac{ d ^{2}y }{dx ^{2} }=\frac{ 2 }{3 }\frac{y*1-x \frac{ dy }{dx } }{ y ^{2} }\] \[\frac{ d ^{2}y }{dx ^{2} }=\frac{ 2 }{3 }\frac{y-x \frac{ 2x }{3y } }{y ^{2} }\] \[\frac{ d ^{2}y}{dx ^{2} }=\frac{ 2 }{ 3 }\frac{ \frac{ 3y ^{2}-2x ^{2} }{ 3y } }{y ^{2} }\] \[\frac{ d ^{2}y }{ dx ^{2} }=-\frac{ 2 }{9 }\frac{ 2x ^{2}-3y ^{2} }{ y ^{3} }\] \[\frac{ d ^{2}y }{ dx ^{2} }=-\frac{ 2 }{9 }\frac{ 4 }{y ^{3} }\] \[\frac{ d ^{2}y }{dx ^{2} }=-\frac{ 8 }{9y ^{3} }\]

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