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Math Help - [SOLVED] Implicit Differentiation

  1. #1
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    [SOLVED] Implicit Differentiation

    y^2=cos(xy)+ln(x+y)

    Find \frac{dy}{dx} in terms of x and y


    Now, this is what I did, I'm not entirely sure if I'm doing it right.

    2y.\frac{dy}{dx} = -ysin(xy).(y+x\frac{dy}{dx})+\frac{1}{x} + \frac{1}{y}\frac{dy}{dx}

    \frac{-1}{x}+ysin(xy)=(\frac{1}{y}-xsin(xy)-2y)\frac{dy}{dx}

    \frac{dy}{dx}=\frac{ysin(xy)-\frac{1}{x}}{\frac{1}{xy}-xsin(xy)-2y}

    So do i get or ?
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  2. #2
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    My working is

    2y*dy=-sin(xy)*(y+x*dy)+(1+dy)/(x+y)
    where dy=dy/dx
    solve dy you'll get the answer?
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Rapid_W View Post
    y^2=cos(xy)+ln(x+y)

    Find \frac{dy}{dx} in terms of x and y


    Now, this is what I did, I'm not entirely sure if I'm doing it right.

    2y.\frac{dy}{dx} = {\color{blue}-ysin(xy).(y+x\frac{dy}{dx})}+{\color{red}\frac{1}{  x} + \frac{1}{y}\frac{dy}{dx}}

    \frac{-1}{x}+ysin(xy)=(\frac{1}{y}-xsin(xy)-2y)\frac{dy}{dx}

    \frac{dy}{dx}=\frac{ysin(xy)-\frac{1}{x}}{\frac{1}{xy}-xsin(xy)-2y}

    So do i get or ?
    can you find your mistake in the blue and the red term or
    want I tell you about them
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  4. #4
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    Not really sure about the blue, but is the red supposed to be

    \frac{1}{xy}\frac{dy}{dx}?
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  5. #5
    MHF Contributor Amer's Avatar
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    ln(x+y)...derive...\frac{1}{x+y}+\frac{dy/dx}{y+x}
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  6. #6
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    ahh ok, should the blue be this?

    -ysin(xy)-xsin(xy).\frac{dy}{dx}
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  7. #7
    MHF Contributor Amer's Avatar
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    yeah that's right you are correct
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  8. #8
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    ok so starting over doing it correctly i get

    2y\frac{dy}{dx} = -sin(xy).y - sin(xy).x\frac{dy}{dx} + \frac{1}{x+y} + \frac{1}{x+y}\frac{dy}{dx}
    <br />
(2y + xsin(xy) - \frac{1}{x+y})\frac{dy}{dx} = -ysin(xy) + \frac{1}{x+y}

    giving me an answer of

    \frac{dy}{dx} = \frac{-ysin(xy)+(x+y)^{-1}}{2y+xsin(xy)-(x+y^{-1}}

    which simplifys to

    \frac{dy}{dx} = \frac{-ysin(xy)}{2y+xsin(xy)} -1

    Is this correct?
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  9. #9
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Rapid_W View Post
    ok so starting over doing it correctly i get

    2y\frac{dy}{dx} = -sin(xy).y - sin(xy).x\frac{dy}{dx} + \frac{1}{x+y} + \frac{1}{x+y}\frac{dy}{dx}
    <br />
(2y + xsin(xy) - \frac{1}{x+y})\frac{dy}{dx} = -ysin(xy) + \frac{1}{x+y}

    giving me an answer of

    \frac{dy}{dx} = \frac{-ysin(xy)+(x+y)^{-1}}{2y+xsin(xy)-(x+y^{-1}}

    which simplifys to

    \frac{dy}{dx} = \frac{-ysin(xy)}{2y+xsin(xy)} -1

    Is this correct?

    \frac{dy}{dx} = \frac{-ysin(xy)+(x+y)^{-1}}{2y+xsin(xy)-(x+y)^{-1}} this is correct

    but how that became to this

    \frac{dy}{dx} = \frac{-ysin(xy)}{2y+xsin(xy)}-1

    I am not sure if it is wrong or not

    \frac{dy}{dx} = \frac{-ysin(xy)+(x+y)^{-1}}{2y+xsin(xy)-(x+y)^{-1}}



    \frac{dy}{dx} =\left(\dfrac{ \dfrac{(x+y)(-ysin(xy))+1 }{(x+y)} }{\dfrac{ (x+y)(2y+xsin(xy)-1)}{(x+y)}}\right)




    \frac{dy}{dx} =\left(\dfrac{ (x+y)(-ysin(xy))+1  }{(x+y)(2y+xsin(xy)-1)}\right)

    simplify it
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  10. #10
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    ahh ok, that makes more sense, ok, my problem is now solved thanks everyone that helped
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