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if $z=x^3 -3xy^2 $ show that

$\frac{\partial^2 z}{\partial x\partial y}=\frac{\partial ^2z}{\partial y\partial z}$

I got $$\frac{\partial^2 z}{\partial x\partial y}=-6y$$

but have no idea how to find $\frac{\partial^2 z}{\partial y\partial z}$ will appreciate if anyone can give any clue..

tharanga dharmapala
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  • $\frac{\partial z^2}{\partial x\partial y}$ or $\frac{\partial^{\color{red}{2}} z}{\partial x\partial y}$ ? – Donald Splutterwit Sep 10 '17 at 10:07
  • $\frac{\partial^{\color{blue}{2}} z}{\partial x\partial y}=\frac{\partial^{\color{blue}{2}} z}{\partial y\partial \color{red}{x}}$ and $\frac{\partial^{\color{blue}{2}} z}{\partial x\partial y}=\color{red}{-}6y$. Correct these & then think about your question again. – Donald Splutterwit Sep 10 '17 at 10:10
  • @DonaldSplutterwit $\frac{\partial^2 z}{\partial x\partial y}$correction thanks.. – tharanga dharmapala Sep 10 '17 at 10:11
  • @DonaldSplutterwit no ,it's said to find $\frac{\partial^2 z}{\partial y\partial z}$ – tharanga dharmapala Sep 10 '17 at 10:21
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    Are you sure ? $\frac{\partial z}{\partial z}=1$ so $\frac{\partial ^2z}{\partial y\partial z}=0$. I think the question wants you to show that it does not matter what order you perform the $x$ and $y$ derivatives. – Donald Splutterwit Sep 10 '17 at 10:25

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