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Math Help - differentiate y = Dsin(kx/y) wrt x

  1. #1
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    Post differentiate y = Dsin(kx/y) wrt x

    Given a function y = Dsin(kx/y) ,where D and k are constants and (kx/y) is in radians
    1. what is dy/dx ?
    2. how can you express x as a function of y , x=?
    3. what is dx/dy ?
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  2. #2
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    You will need to use the chain rule and implicit differentiation...

    y = D\sin{\frac{kx}{y}}

    Let u = \frac{kx}{y} so that y= D\sin{u}.


    \frac{dy}{du} = D\cos{u} = D\cos{\left(\frac{kx}{y}\right)}.


    \frac{du}{dx} = \frac{d}{dx}\left(\frac{kx}{y}\right)

     = k\,\frac{d}{dx}\left(\frac{x}{y}\right)

     = k\left(\frac{y\,\frac{d}{dx}(x) - x\,\frac{d}{dx}(y)}{y^2}\right)

     = \frac{k\left(y - x\,\frac{dy}{dx}\right)}{y^2}.


    So \frac{dy}{dx} = \frac{dy}{du}\,\frac{du}{dx}

    \frac{dy}{dx} = \frac{Dk\left(y - x\,\frac{dy}{dx}\right)\cos{\left(\frac{kx}{y}\rig  ht)}}{y^2}

    y^2\,\frac{dy}{dx} = Dky\cos{\left(\frac{kx}{y}\right)} - Dkx\cos{\left(\frac{kx}{y}\right)}\frac{dy}{dx}

    y^2\,\frac{dy}{dx} + Dkx\cos{\left(\frac{kx}{y}\right)}\frac{dy}{dx} = Dky\cos{\left(\frac{kx}{y}\right)}

    \frac{dy}{dx}\left[y^2 + Dkx\cos{\left(\frac{kx}{y}\right)}\right] = Dky\cos{\left(\frac{kx}{y}\right)}

    \frac{dy}{dx} = \frac{Dky\cos{\left(\frac{kx}{y}\right)}}{y^2 + Dkx\cos{\left(\frac{kx}{y}\right)}}
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    Thanks

    Thanks for the assistance, much appreciated.
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