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Okay the title makes no sense.

I have a two variable function, $f(x,t)$. When is it that $$ \left(\frac \partial{\partial x} f(x,t) \right)\bigg| _{t=0} = \frac{d}{dx} f(x,0)$$?

My guess is that it works if $f(x,t)$ is continuous with respect to $t$ in $0$, but I'm not sure.

Note that it does not work in general,since $f(x,t) = x^t / t$ is a counterexample;

$$\frac 1x = \left(\frac \partial{\partial x} f(x,t) \right)\bigg| _{t=0} \neq \frac{d}{dx} f(x,0) $$

As $f(x,0)$ does not exists in the first place.

Ant
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  • Basically your question demands to know when can we interchange the limits in the expression $$\lim_{t\to 0}\lim_{\delta x\to 0}\frac{f(x+\delta x,t)-f(x,t)}{\delta x}$$, right? – Samrat Mukhopadhyay Jul 20 '15 at 11:00
  • I think you can find relevant information for solving this question in the answer to this question. – Samrat Mukhopadhyay Jul 20 '15 at 11:07
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    By definition $\frac{\partial}{\partial x}f(x,0)=\lim_{h\to 0}\frac{f(x+h,0)-f(x,0)}{h}$, no limit wrt $t$ is involved. The $t$-limit appears if you want your derivative to be continuous in $t$. Note that $x^t/t$ is not defined for $t=0$, it makes no sense to talk about $\frac{\partial}{\partial x}f(x,0)$. – A.Γ. Jul 20 '15 at 11:25
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    like @A.G. points out, the equation is the very definition of partial derivative. Your example is just a non-example. – user251257 Jul 20 '15 at 12:33
  • @user251257 What? What do you mean the equation is the very definition of partial derivative? I have $$\lim_{h \to 0} \frac{f(x + h, t) - f(x,t)}{h} \bigg|{t = 0} = \lim{h \to 0} \frac{f(x+h, 0) - f(x, 0)}{h}$$. Are they always equal? I don't think so, because as my example show, the LHS can exists while the RHS does not. That was the only point of the example. Under which conditions then it's true? Really, I have the feeling I am not being understood, which is probably my fault, but try to be more clear about what's wrong with my reasoning – Ant Jul 20 '15 at 12:37
  • What do you think that $|_{t=0}$ means? – user251257 Jul 20 '15 at 12:37
  • @user251257 Calculate the limit in $t=0$. That limit defines a function $g(x,t)$ and I want to calculate that in $t=0$. Note that I can't just put inside the limit $t=0$ (to have trivially equality with RHS) because then (as my example shows) I'd have something that does not exists, while the LHS is actually defined. If I'm wrong then please explain to me what $\mid_{t=0}$ means – Ant Jul 20 '15 at 12:40
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    You can and you should. In your example, $f(x,0)$ does not exist, let alone $\partial_x f(x,0)$. You haven't compute $\partial_x f(x,0)$ but rather $\partial_x f(x,t)$ for $t\ne 0$ and want to extend it some how to $\partial_x f(x,0)$, which is not valid in general. – user251257 Jul 20 '15 at 12:43
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    @Ant $|_{t=0}$ means exactly "set $t$ to zero". $x$ and $t$ are independent variables for $f$, and the calculation of $x$-derivative is not interfered with $t$, it is calculated pointwise wrt $t$ (by definition of a partial derivative). – A.Γ. Jul 20 '15 at 13:35

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