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if $u(x,t)$ differentiable function and i only have $u(x,0)$, then is it right $\frac{\partial(x,0)}{\partial x} = \left.\frac{\partial(x,t)}{\partial x}\right|_{t=0}$ or can i derive $u(x,0)$ to $x$ without to have to know exactly $u(x,t)$?

What the counterexample if it's wrong and what's the prove or the theorem to use if it's right?

Thank you for attention.

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$$\left.\frac{\partial f(x,t)}{\partial x}\right|_{t=0}=\left.\lim_{h\to 0}\frac{f(x+h,t)-f(x,t)}{h}\right|_{t=0}=\lim_{h\to 0}\frac{f(x+h,0)-f(x,0)}{h}=\frac{\partial f(x,0)}{\partial x}$$

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  • $\begingroup$ Thank you Dr. MV. it's back to basic. $\endgroup$ – Wily Jun 8 '15 at 5:42
  • $\begingroup$ You're welcome. My pleasure. $\endgroup$ – Mark Viola Jun 8 '15 at 13:21
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Yes, you can interchange plugging in $t=0$ and differentiating with respect to $x$ (although it is important here that the value you are plugging in does not depend on $x$).

You can formalize this by writing down a function $g(x) = (x,0)$ and differentiating $f\circ g$. By the chain rule you get

\begin{align*} \frac{\partial f\circ g}{\partial x} &= D_1 f[g(x)] \frac{\partial g_1}{\partial x} + D_2 f[g(x)] \frac{\partial g_2}{\partial x}\\ &= D_1f[g(x)]\cdot 1 + 0\\ &= \frac{\partial f}{\partial x}(x,0), \end{align*} where $D_i$ is differentiation with respect to the $i$th parameter.

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  • $\begingroup$ Thank you. But what does $g_1 \& g_2$ mean ? $\endgroup$ – Wily Jun 8 '15 at 5:51
  • $\begingroup$ @Wily First and second coordinate of $g$. $\endgroup$ – user7530 Jun 8 '15 at 7:15

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