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Math Help - Finding dy/dx by implicit differentiation

  1. #1
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    Finding dy/dx by implicit differentiation

    How do I find dy/dx of this problem?

    tan (x/y) = x + y
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by !!! View Post
    How do I find dy/dx of this problem?

    tan (x/y) = x + y
    do you realize that everytime you differentiate a y-term you attach dy/dx to it?

    so, to start you off.

    \tan \frac xy = x + y

    \Rightarrow \sec^2 \left( \frac xy \right) \cdot \frac d{dx} \left( \frac xy \right) = 1 + \frac {dy}{dx}

    (the left hand side follows by the chain rule)

    can you continue?
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  3. #3
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    I actually forgot what I did, redid the problem and got the answer correct, but thank you!

    Can you help me with another one, please?
    xy = cot (xy)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by !!! View Post
    I actually forgot what I did, redid the problem and got the answer correct, but thank you!

    Can you help me with another one, please?
    xy = cot (xy)
    use the product rule on the left and the chain rule (coupled with the product rule) on the right

    to start you off.

    xy = \cot (xy)

    \Rightarrow y + x~\frac {dy}{dx} = - \csc^2 (xy) \cdot \frac d{dx} [xy]

    can you continue?
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