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sha0403

  • 2 years ago

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

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  1. winterfez
    • 2 years ago
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    \[2yy'=y'e ^{x^2}+2\] \[2yy'=y'e ^{x^2}+y2xe ^{x^2}+2\] \[2yy'-y'e ^{x^2}=y2xe ^{x^2}+2\] \[y'(2y-e ^{x^2})=y2xe ^{x^2}+2\] \[y'=\frac{ y2xe ^{x^2}+2 }{ 2y-e ^{x^2} }\]

  2. winterfez
    • 2 years ago
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    I'm assuming \[\frac{ dy^2 }{ dx^2 }\] is second derivative?

  3. sha0403
    • 2 years ago
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    thanks,, but to differentiate the answer again i have no idea..

  4. winterfez
    • 2 years ago
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    k let me try...it going to be long

  5. sha0403
    • 2 years ago
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    ok tq so much

  6. winterfez
    • 2 years ago
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    \[y''=\frac{ (2y'xe ^{x^2}+2ye ^{x^2}+2xy(2x)e ^{x^2})(2y-e ^{x^2})-(y2xe ^{x^2}+2)(2y'-2xe ^{x^2} }{ (2y-e ^{x^2})^2 }\] \[\frac{ (2y'xe ^{x^2}+2ye ^{x^2}+4x^2ye ^{x^2})(2y-e ^{x^2})-(y2xe ^{x^2}+2)(2y'-2xe ^{x^2}) }{ (2y-e ^{x^2})^2 }\]

  7. winterfez
    • 2 years ago
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    phewww this is what I got so far

  8. sha0403
    • 2 years ago
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    the answer just like that?

  9. sha0403
    • 2 years ago
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    ok tq so much..now i get it

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