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anonymous

  • 5 years ago

Find dy/dx by implicit differentiation" xe^y-10x+3y=0 i really do need help!

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  1. amistre64
    • 5 years ago
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    implicit differentation means you assume y is a function of x: Dx(xe^y) -Dx(10x) +Dx(3y) = Dx(0) Dx(xe^y) use the product rule: (dx/dx)e^y + (dy/dx)xe^y = e^y + y'xe^y Dx(10x) = (dx/dx)(10) = 10 Dx(3y) = 3(dy/dx) = 3y' Dx(0) = (d0/dx)0 = 0 Get all your y' on one side and everything else to the other: e^y + y'xe^y -10 +3y' = 0 y'xe^y + 3y' = 10 -e^y y'(xe^y +3) = 10 -e^y (dy/dx) =y' =(10-e^y) / (xe^y +3) That should be it :)

  2. anonymous
    • 5 years ago
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    omg!! thank you so much!!!

  3. amistre64
    • 5 years ago
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    Implicits confused me at first; but they are just like ll the rest :)

  4. anonymous
    • 5 years ago
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    im a lil confused, but im gonna keep studying thank you again

  5. anonymous
    • 5 years ago
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    gratz

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