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Math Help - Implicit differentiation

  1. #1
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    Implicit differentiation

    I'm having trouble getting:

    ln(x/y)=(e^-x)(cos(y/2))

    down to where I can separate y'. I was able to get it down to:

    (1/x)-(1/y)(y')=e^-x (sin(y/2)(y'/2))

    d/dx cos(y/2)= -sin(y/2)(y'/2) correct?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by ppark View Post
    ln(x/y)=(e^-x)(cos(y/2))

    down to where I can separate y'. I was able to get it down to:

    (1/x)-(1/y)(y')=e^-x (sin(y/2)(y'/2))

    Check the right side:

    -e^{-x}\cos (y/2)+\ldots
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by ppark View Post
    I'm having trouble getting:

    ln(x/y)=(e^-x)(cos(y/2))

    down to where I can separate y'. I was able to get it down to:

    (1/x)-(1/y)(y')=e^-x (sin(y/2)(y'/2))

    d/dx cos(y/2)= -sin(y/2)(y'/2) correct?
    The derivative of the left hand side is:

    \displaystyle \frac{d}{dx}\left( \ln(x/y)\right)=\frac{y}{x}\frac{d}{dx}\left( \frac{x}{y}\right)=\frac{y}{x}\left(\frac{1}{y}-\frac{x}{y^2}y' \right)= \ ...

    so that looks OK.

    Now the derivative of the right hand side:

    \displaystyle \frac{d}{dx}\left( e^{-x}\cos(y/2)\right)=-e^{-x}\cos(y/2)-e^{-x}\sin(y/2)\frac{y'}{2}=\ ...

    CB
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    The derivative of the left hand side is:

    \displaystyle \frac{d}{dx}\left( \ln(x/y)\right)=\frac{y}{x}\frac{d}{dx}\left( \frac{x}{y}\right)=\frac{y}{x}\left(\frac{1}{y}-\frac{x}{y^2}y' \right)= \ ...

    so that looks OK.

    Now the derivative of the right hand side:

    \displaystyle \frac{d}{dx}\left( e^{-x}\cos(y/2)\right)=-e^{-x}\cos(y/2)-e^{-x}\sin(y/2)\frac{y'}{2}=\ ...

    CB
    Wow thank you, can't believe I forgot to apply the product rule...

    actually I got this the first time I tried the problem, but the problem with doing it this way is when I get all the y' on one side there doesn't seem a way to take out the y'.

    ((2/y')-(y'/y))=(((-e^-x)(cos(y/2)))-(e^-x)(sin(y/2)-(1/x))

    ((2y-(y'^2))/(y'(y)))=...
    Last edited by ppark; April 9th 2011 at 08:31 AM.
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