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Find partial derivatives of $f(g(x,y))$ where $f(z)=cos(z)$ and $g(x,y)=\frac xy$

I understand the principal of chain rule for multivariable calculus, however this example confuses me. As I would approach it, is to write it out in full as following: $$\frac{df(g(x,y))}{dx}=\frac{df}{dg}\frac{dg}{dx}$$ and $$\frac{df(g(x,y))}{dy}=\frac{df}{dg}\frac{dg}{dy}$$

Easy to find that $\frac{dg}{dx}=\frac 1x$ and $\frac{dg}{dy}=-\frac xy^2$ However $\frac{df}{dg}$ confuses me, as I'm not sure what to do when I need to differentiate with respect to the variable, which is not in a function.

user140172
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  • So the function you are differentiating here is $\cos(\frac{x}{y})$. Can you identify how to apply the chain rule to each part of this function now? – Autolatry Apr 03 '14 at 16:21
  • $D(fog) = Df\cdot Dg = -sin(z)|_{z=g(x,y)} \cdot (1/y, -x/y^2) = (\frac{-sin(x/y)}{y}, \frac{-xsin(x/y)}{y^2})$... – Christopher K Apr 03 '14 at 16:23

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