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anonymous

  • one year ago

calculate dy/dx if Ln(x+y)=e^x/y

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  1. welshfella
    • one year ago
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    this is implicit differentiation. Treat y as a function of x

  2. anonymous
    • one year ago
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    can you walk me through solving problem

  3. welshfella
    • one year ago
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    give me a minute - I haven't done these for a while...

  4. welshfella
    • one year ago
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    you differentiate term by term so first do ln(x + y) you use the chain rule here as you have a function within a function are you familiar with the chain rule?

  5. anonymous
    • one year ago
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    yes

  6. welshfella
    • one year ago
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    ok so derivative of ln (x + y) = 1 / (x + y) * d(x+y)/dx = 1( x + y) * ( 1 + y') = 1 (x +y) + y' / (x + y) ( I've written dy/dx as y')

  7. welshfella
    • one year ago
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    that last line is 1 / (x + y) + y' / ( x + y)

  8. anonymous
    • one year ago
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    ok

  9. welshfella
    • one year ago
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    now find derivative of the right hand side Use the quotient rule =[ y* e^x - e^x * y'] / y^2 so we have 1 / (x + y) + y' / ( x + y) = [ e^x( y - y')] / y^2 now you need to solve for y'

  10. anonymous
    • one year ago
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    ok

  11. welshfella
    • one year ago
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    its a bit messy first this i would do is multiply thru by y^2( x + y)

  12. anonymous
    • one year ago
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    yes this is allot

  13. welshfella
    • one year ago
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    I got dy/dx = xy e^x + y^2 e^x - y^2 ------------------- y^2 + x e^x + y e^x but work it but for yourself

  14. anonymous
    • one year ago
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    thanks this sight is super helpful and thank you so much

  15. welshfella
    • one year ago
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    yw

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