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anonymous

  • 5 years ago

use implicit differentiation to find dy/dx - 2xy-y^2= 1 can someone explain step by step with formulas explaining in detail.

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  1. bahrom7893
    • 5 years ago
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    Let me call dy/dx y'

  2. bahrom7893
    • 5 years ago
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    use product rule for -2xy and power rule for -y^2: -2x*y' + y*(-2) - 2y*y' = 0

  3. bahrom7893
    • 5 years ago
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    -2xy' - 2y - 2yy' = 0, divide everything by -2: xy' + y + yy' = 0

  4. bahrom7893
    • 5 years ago
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    now move y to the right side: xy' + yy' = -y

  5. anonymous
    • 5 years ago
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    you lost the =1

  6. bahrom7893
    • 5 years ago
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    where?

  7. anonymous
    • 5 years ago
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    -2xy-y^2 = 1 is original problem right?

  8. bahrom7893
    • 5 years ago
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    yeah, but derivative of a constant is 0!

  9. anonymous
    • 5 years ago
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    gotcha

  10. anonymous
    • 5 years ago
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    what is the formula for implicit differentiation ?

  11. bahrom7893
    • 5 years ago
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    now move y to the right side: xy' + yy' = -y take y' out: y'(x+y) = -y

  12. bahrom7893
    • 5 years ago
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    there is no formula, u use a combination of several formulas

  13. anonymous
    • 5 years ago
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    oh ah...ok let me look and little by little look into what u wrote to understand

  14. bahrom7893
    • 5 years ago
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    y' = dy/dx = -y/(x+y). Fan me if I helped, thanx! =)

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