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Math Help - Differentiating sin(x/y)

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    Differentiating sin(x/y)

    How do we differentiate something like this? Chain rule should be used here, so the first term will be cos(x/y), but then I don't understand how to break up this (x/y) function?
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    Re: Differentiating sin(x/y)

    Oh, wait, I just looked up the answer. It seems like I should just use quotient rule?
    Last edited by maxpancho; August 9th 2014 at 07:49 AM.
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    Re: Differentiating sin(x/y)

    First, the answer is yes. By the way, it is d/dx right? What you are differentiating WITH RESPECT TO is important.
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    Re: Differentiating sin(x/y)

    Yes d/dx.
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    Re: Differentiating sin(x/y)

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    Re: Differentiating sin(x/y)

    *deleted*
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    Re: Differentiating sin(x/y)

    Are we assuming that y is a function of x? Or are you doing a partial derivative?
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    Re: Differentiating sin(x/y)

    It's actually an expression in implicit differentiation.
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    Re: Differentiating sin(x/y)

    In other words, y is a function of x.

    So like you have suggested, you need to differentiate this function implicitly. How will you do this?
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    Re: Differentiating sin(x/y)

    $\cos(\dfrac{x}{y}) \cdot \dfrac{y-xy'}{y^2}$ right?
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    Re: Differentiating sin(x/y)

    That's correct for now. But remember, our aim is to solve for y' so that means your original expression needs to be an EQUATION.

    What was the original equation?
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    Re: Differentiating sin(x/y)

    Well, I just wanted to clear up this part , that's why I posted only it. I was confused whether I should apply the chain rule to the inside term x\y and how if so.
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    Re: Differentiating sin(x/y)

    You already did. But like I said, you can not continue unless you have an EQUATION to start with.
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    Re: Differentiating sin(x/y)

    $$\frac{d}{dx}\left[\sin(\dfrac{x}{y})\right]=\frac{d}{dx}\left[ x+y \right]$$

    $$\cos(\dfrac{x}{y}) \cdot \dfrac{y-xy'}{y^2}=1+y'$$

    $$\cos(\dfrac{x}{y})y - \cos(\dfrac{x}{y})xy' = y^2+y^2y'$$

    $$\cos(\dfrac{x}{y})y - y^2 = \cos(\dfrac{x}{y})xy' + y^2y'$$

    $$\cos(\dfrac{x}{y})y - y^2 = y' \left( \cos(\dfrac{x}{y})x + y^2 \right) $$

    $$y'=\dfrac{\cos(\dfrac{x}{y})y - y^2}{\cos(\dfrac{x}{y})x + y^2 }$$
    Last edited by maxpancho; August 10th 2014 at 04:46 AM.
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    Re: Differentiating sin(x/y)

    \frac{dy}{dx} = \frac{-f_{x}}{f_{y}} where f_{x}, f_{y} are the partial derivatives of x and y respectively. If you don't know what partial derivatives are, then for the x partial. treat y as though it is a constant. and similarly for the y partial, treat x as it is a constant.


    I get  \frac{dy}{dx} = \frac{y}{x}
    Last edited by Jonroberts74; August 10th 2014 at 12:07 PM.
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