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Let $ w=\frac{x^2}{y} $ with $ x=e^{-u^2}u $ and $ y=e^{-u^2}v $


$ \frac{\partial w}{\partial x} = (x^2)'\frac{1}{y} = \frac{2x}{y} $

$ \frac{\partial w}{\partial y} = x^2(\frac{1}{y})' = -\frac{x^2}{y^2} $

$ \frac{\partial y}{\partial u} = (e^{-u^2})'v = e^{-u^2}(-u^2)'v = -2e^{-u^2}uv $

$ \frac{\partial y}{\partial v} = e^{-u^2}(v)' = e^{-u^2} $

$ \frac{\partial x}{\partial u} = (e^{-u^2}u)' = (e^{-u^2})'u+u'e^{-u^2} = e^{-u^2}(-u^2)'u+e^{-u^2} = -2e^{-u^2}u^2 + e^{-u^2} = e^{-u^2}(1-2u^2) $

$ \frac{\partial x}{\partial v} = 0 $


$ \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} = \frac{2x}{y}e^{-u^2}(1-2u^2) +(-\frac{x^2}{y^2})(-2e^{-u^2}uv) = \frac{2e^{-u^2}u}{e^{-u^2}v}e^{-u^2}(1-2u^2) + 2e^{-u^2}uv\frac{(e^{-u^2}u)^2}{(e^{-u^2}v)^2} = \frac{2e^{-u^2}u}{v}-\frac{4e^{-u^2}u^3}{v} + \frac{2e^{-u^2}u^3}{v} = \frac{2e^{-u^2}u}{v} - \frac{2e^{-u^2}u^3}{v} = \frac{2e^{-u^2}u}{v}(1-u^2) $

$ \frac{\partial w}{\partial v} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial v} = - \frac{x^2}{y^2}e^{-u^2} = - \frac{(e^{-u^2}u)^2}{(e^{-u^2}v)^2} e^{-u^2} = - \frac{e^{-u^2}u^2}{v^2} $

Is my solution correct? I am not sure if I understood chain rule according to partial derivatives.

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up vote 3 down vote accepted

Yes you get it. Everything is correct, but be careful when you write $f'$: this notation is reserved for derivative of a function depending only on one variable.

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Thank you for your quick response! $ f' $ used just for shortness. – Edward Ruchevits Aug 29 '12 at 19:15

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