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Arctan(y)=3x+y how would I go about finding y'?

  1. Feb 11, 2009 #1
    1. The problem statement, all variables and given/known data

    Arctan(y)=3x+y how would I go about finding y'?

    2. Relevant equations



    3. The attempt at a solution

    I tried to start out with...

    arctan(y)= tan(x)
    tan(x)=3x+y
    tan(x)-3x=y

    y'=1/(1+x^2) -3

    is this correct?
    thanks in advance.
     
  2. jcsd
  3. Feb 11, 2009 #2
    Re: Arctan(y)=3x+y

    How you reached here from
    Arctan(y)=3x+y ?

    What's
    d/dx (Arctan(y)) ?
    Use the chain rule.
     
  4. Feb 11, 2009 #3

    Dick

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    Re: Arctan(y)=3x+y

    How can arctan(y)=tan(x) when your equation says arctan(y)=3x+y?? That's not right. Don't try to solve for y. You can't. Differentiate it implicitly.
     
  5. Feb 11, 2009 #4
    Re: Arctan(y)=3x+y

    d/dx arctan y

    y'/1+y^2*y' ?
     
  6. Feb 11, 2009 #5

    Dick

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    Re: Arctan(y)=3x+y

    Close. But why are there two y' in there? And use parentheses to avoid confusion. There's a difference between 1/1+y^2 and 1/(1+y^2).
     
  7. Feb 11, 2009 #6
    Re: Arctan(y)=3x+y

    y'/(1+y^2) *y'
    I thought the 2nd y' has to be there bc for example
    y=arccot(x^2)
    x^2=cot y
    2x=-csc^2(y) *y'

    is that not the case since I already have a y'?
     
  8. Feb 11, 2009 #7
    Re: Arctan(y)=3x+y

    oops nvm...
     
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