1. ## 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?

2. ## Re: Differentiating sin(x/y)

Oh, wait, I just looked up the answer. It seems like I should just use quotient rule?

3. ## 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.

Yes d/dx.

*deleted*

7. ## Re: Differentiating sin(x/y)

Are we assuming that y is a function of x? Or are you doing a partial derivative?

8. ## Re: Differentiating sin(x/y)

It's actually an expression in implicit differentiation.

9. ## 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?

10. ## Re: Differentiating sin(x/y)

$\cos(\dfrac{x}{y}) \cdot \dfrac{y-xy'}{y^2}$ right?

11. ## 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?

12. ## 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.

13. ## Re: Differentiating sin(x/y)

You already did. But like I said, you can not continue unless you have an EQUATION to start with.

14. ## Re: Differentiating sin(x/y)

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

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

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

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

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

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

15. ## Re: Differentiating sin(x/y)

$\displaystyle \frac{dy}{dx} = \frac{-f_{x}}{f_{y}}$ where $\displaystyle f_{x}, f_{y}$ 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 $\displaystyle \frac{dy}{dx} = \frac{y}{x}$

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