anonymous
  • anonymous
can u all help me to differentiate this, i try but still did not get it... find dy/dx and d^2y/dx^2question: y^2=ye^(x^2)+2x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[2yy'=y'e ^{x^2}+2\] \[2yy'=y'e ^{x^2}+y2xe ^{x^2}+2\] \[2yy'-y'e ^{x^2}=y2xe ^{x^2}+2\] \[y'(2y-e ^{x^2})=y2xe ^{x^2}+2\] \[y'=\frac{ y2xe ^{x^2}+2 }{ 2y-e ^{x^2} }\]
anonymous
  • anonymous
I'm assuming \[\frac{ dy^2 }{ dx^2 }\] is second derivative?
anonymous
  • anonymous
thanks,, but to differentiate the answer again i have no idea..

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anonymous
  • anonymous
k let me try...it going to be long
anonymous
  • anonymous
ok tq so much
anonymous
  • anonymous
\[y''=\frac{ (2y'xe ^{x^2}+2ye ^{x^2}+2xy(2x)e ^{x^2})(2y-e ^{x^2})-(y2xe ^{x^2}+2)(2y'-2xe ^{x^2} }{ (2y-e ^{x^2})^2 }\] \[\frac{ (2y'xe ^{x^2}+2ye ^{x^2}+4x^2ye ^{x^2})(2y-e ^{x^2})-(y2xe ^{x^2}+2)(2y'-2xe ^{x^2}) }{ (2y-e ^{x^2})^2 }\]
anonymous
  • anonymous
phewww this is what I got so far
anonymous
  • anonymous
the answer just like that?
anonymous
  • anonymous
ok tq so much..now i get it

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