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Let $u(x,y)=f(x+y)+g(x-y)$. How can I calculate $\partial u/ \partial x$, $\partial u/ \partial y$, $\partial^2 u/ \partial x^2$, $\partial^2 u/ \partial y^2$ in terms of derivatives of $f$ and $g$.

So I thought of making a new variable such that say, $v=x+y, w=x-y$ and then substituting. But I don’t know where this will lead to. Can someone help me on this path?

sophie
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1 Answers1

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The two rules you need are

$$\def\p#1#2{\frac{\partial #1}{\partial #2}}\p{(f+g)}x=\p fx+\p gx$$

and

$$\p{f(g(x,y))}x=f'(g(x,y))\p{g(x,y)}x\;.$$

joriki
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  • Can you just put the answers so I can check? – sophie Jul 09 '15 at 20:43
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    No. But you can use this to get the answers and then I can check. – joriki Jul 09 '15 at 20:43
  • for partial du/dx is it (1+y)f'(v)+(1-y)f'(w)? – sophie Jul 09 '15 at 20:46
  • @sophie: Almost. Since only $x$ and $y$ appear in the problem statement, we can safely assume that the intended meaning of the ambiguous partial derivative operators is that $x$ and $y$ are being considered as two independent variables and $\partial/\partial x$ refers to the variation with $x$ at constant $y$. What is the variation of $y$ at constant $y$? – joriki Jul 09 '15 at 20:50
  • Is it 1? Or is that wrong? – sophie Jul 09 '15 at 20:50
  • @sophie: How can something constant vary? Think of a function of $x$ and $y$ as a landscape plotted over the $x$-$y$-plane. The function $h(x,y)=y$ rises linearly in the $y$ direction and is constant in the $x$ direction. The partial derivative $\partial/\partial x$ measures the rate of variation in the $x$ direction, at constant $y$, and in this direction $h(x,y)=y$ is constant, independent of $x$. What is the derivative of a constant independent of $x$ with respect to $x$? – joriki Jul 09 '15 at 21:20