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Math Help - Implicit differentiation with quotients

  1. #1
    Member Chokfull's Avatar
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    Implicit differentiation with quotients

    Find \frac {dy} {dx} by implicit differentiation.

     <br />
\tan \frac {x} {y}=x+y<br />

    By the chain rule, the left side becomes

    \sec^2 (\frac {x} {y}) * (the derivative of \frac {x} {y})

    I have two problems, one, where do the y-primes go, and two, how do I solve for it?

    The final answer should look like

    y'= \frac {y \sec^2 \frac {x} {y} - y^2} {y^2 + x \sec^2 \frac {x} {y}}
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by Chokfull View Post
    Find \frac {dy} {dx} by implicit differentiation.

     <br />
\tan \frac {x} {y}=x+y<br />

    By the chain rule, the left side becomes

    \sec^2 (\frac {x} {y}) * (the derivative of \frac {x} {y})

    I have two problems, one, where do the y-primes go, and two, how do I solve for it?

    The final answer should look like

    y'= \frac {y \sec^2 \frac {x} {y} - y^2} {y^2 + x \sec^2 \frac {x} {y}}

    \sec^2 (\frac {x} {y}) \frac{y-xy'}{y^2} = 1 + y'

    Now we need to expand and group y' terms on one side. I'll try to move the y' terms on right hand side:

    \sec^2 (\frac {x} {y}) (y-xy') = y^2 + y^2y'

    Expand on left side:

    y \sec^2 (\frac {x} {y})-x\sec^2 (\frac {x} {y})y' = y^2 + y^2y'

    Move y' terms on the right and y^2 from right to left:

    y \sec^2 (\frac {x} {y}) - y^2 = y^2y' +x\sec^2 (\frac {x} {y})y'

    Factor a y', then divide to solve for it. I hope my algebra does not contain any errors above.

    Good luck!
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  3. #3
    Member Chokfull's Avatar
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    Quote Originally Posted by apcalculus View Post
    \sec^2 (\frac {x} {y}) \frac{y-xy'}{y^2} = 1 + y'

    Now we need to expand and group y' terms on one side. I'll try to move the y' terms on right hand side:

    \sec^2 (\frac {x} {y}) (y-xy') = y^2 + y^2y'

    Expand on left side:

    y \sec^2 (\frac {x} {y})-x\sec^2 (\frac {x} {y})y' = y^2 + y^2y'

    Move y' terms on the right and y^2 from right to left:

    y \sec^2 (\frac {x} {y}) - y^2 = y^2y' +x\sec^2 (\frac {x} {y})y'

    Factor a y', then divide to solve for it. I hope my algebra does not contain any errors above.

    Good luck!

    OK thanks a lot! It looks like i had the derivative right, then I was mainly going the wrong way in my algebra
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