# How do I differentiate this?

I am given $u=\frac{x+y}{\sqrt 2}$ and $v=\frac{x-y}{\sqrt2}$, how would I find $\frac{d^2}{du^2},\frac{d^2}{dv^2}$?

I rearranged $u$ and $v$ in terms of $x$ and $y$, and I get $x = \frac{u + v}{\sqrt{2}}$ and $y = \frac{u - v}{\sqrt{2}}$. But how do I find $\frac{d^2}{du^2}$ and $\frac{d^2}{dv^2}$?

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Thanks, so is the answer for both, $0$? –  Ray Jan 16 '12 at 19:51
What are you differentiating in this expression: $\frac{d^2}{dv^2}$? –  Emmad Kareem Jan 16 '12 at 19:52
@Emmad: It doesnt state anything in the problem, although you can take a look at the original question below (note the $v$ changes to $w$ in the original question). math.stackexchange.com/questions/99608/… –  Ray Jan 16 '12 at 19:55
sorry this is not too advanced for me :) –  Emmad Kareem Jan 16 '12 at 20:05
haha no problem, its ok. –  Ray Jan 16 '12 at 20:13

if $f$ is a function of $x,y$ then $$\frac{\partial f}{\partial u}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}$$ and $$\frac{\partial^2 f}{\partial u^2}=\frac{\partial}{\partial u}\Big(\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}\Big)$$ $$=\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial u^2}+\frac{\partial x}{\partial u}\Big(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial u} +\frac{\partial^2 f}{\partial y\partial x}\frac{\partial y}{\partial u}\Big) +\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial u^2}+\frac{\partial y}{\partial u}\Big(\frac{\partial^2 f}{\partial x\partial y}\frac{\partial x}{\partial u}+\frac{\partial^2 f}{\partial y^2}\frac{\partial y}{\partial u}\Big)$$ $$=\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial u^2}+\frac{\partial f}{\partial y}\frac{\partial^2 y}{\partial u^2}+2\frac{\partial^2 f}{\partial x\partial y}\frac{\partial x}{\partial u}\frac{\partial y}{\partial u}+\Big(\frac{\partial x}{\partial u}\Big)^2\frac{\partial^2 f}{\partial x^2}+\Big(\frac{\partial y}{\partial u}\Big)^2\frac{\partial^2 f}{\partial y^2}$$ note $$x=(u+v)\sqrt{2}, y=(u-v)\sqrt{2},\frac{\partial x}{\partial u}=\frac{\partial y}{\partial u}=\sqrt{2}$$ so the above simplifies to $$4\frac{\partial^2 f}{\partial x\partial y}+2\Big(\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}\Big)$$ you can do the same thing with respect to $v$ (i assumed equality of mixed partials above)

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Thanks, Im just wondering, how did you get $x=(u+v)\sqrt{2}$ and $y=(u-v)\sqrt{2}$? –  Ray Jan 16 '12 at 21:08

When you apply $\partial_u$ you take the directional derivative with respect to the vector $\frac 1{\sqrt 2}(1,1)$, so $\partial u =\frac 1{\sqrt 2}(\partial x+\partial y)$. Now take the directional derivative with respect to the vector $\frac 1{\sqrt 2}(1,1)$ of $\partial u$. We get $\partial^2_u=\frac 12(\partial^2_x+\partial_x\partial_y+\partial_y\partial_x+\partial_y^2)$. Now, do the same for $\partial_v$ and $\partial_v^2$.

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Thanks, I understood what you did, but for $\partial^2u=\frac 12(\partial^2_x+\partial_x\partial y+\partial_y\partial_x+\partial_y^2)$, what exactly do I differentiate? What would $\partial^2_x$ be for example? –  Ray Jan 16 '12 at 21:12
$\partial_x^2$ apply to a function $f$ is only the the two times derivative with respect to $x$. –  Davide Giraudo Jan 16 '12 at 21:14
Thanks, but what would the $f$ be? $\frac{x+y}{\sqrt 2}$? –  Ray Jan 16 '12 at 21:22
Not necessary, here $f$ is only a function on which you apply the operator $\partial_u$. –  Davide Giraudo Jan 16 '12 at 21:27
Thanks, Im sorry but im really stuck on this question. How would I find $\partial^2_x,\partial_x\partial_y,\partial_y\partial_x,\partial_y^2$ etc.? As in, what do I differentiate when I am trying to find $\partial_x^2$? –  Ray Jan 16 '12 at 21:40