Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
Thank you for your quick response! $ f' $ used just for shortness. –  Edward Ruchevits Aug 29 '12 at 19:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.