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I'm trying to calculate a Christoffel symbol, but I'm stuck on showing that $$\frac{\partial(e^{2A})}{\partial r}=2e^{2A}\frac{\partial A}{\partial r}\ ? $$

Nice easy steps would be much appreciated.

Thank you

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    What exactly is the problem? This is the chain rule. Assumming $A = A(r)$, $A$ is differentiable w.r. to $r$ and $e^x$ is the common exponential function.2011-12-05
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    Thomas, a bit simpler please. What form of chain rule? I know $\frac{d\left(e^{x}\right)}{dx}=e^{x}$ but how do I calculate $\frac{\partial\left(e^{2A}\right)}{\partial r}$ . Yes, $A$ is a function of $r$ . Afraid I've never heard of “common exponential function”.2011-12-05
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    It's just the ordinary chain rule. Would you know how to differentiate $e^{2r^5}$, for example?2011-12-05
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    Hans Lundmark - I'd let $u=2r^{5}$, $\frac{du}{dr}=10r^{4},y=e^{u}$ $\frac{dy}{du}=ue^{u}$ (I think), therefore $\frac{dy}{dr}=\frac{du}{dr}\frac{dy}{du}$ . Is that OK? I think I'm confused about $A$ not being an explicit function of $r$ . Thanks.2011-12-05
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    @Peter4075: Good! Then you've got it! Just take $u=2A$ in your case. You will have $du/dr=2 dA/dr$, and that's all there is to it.2011-12-05
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    Hans Lundmark: $u=A$ gives $\frac{du}{dr}=2\frac{dA}{dr}$ (though I'd normally want to differentiate wrt A, but I'll take your word that's OK), then $\frac{d\left(e^{2A}\right)}{du}=\frac{d\left(e^{u}\right)}{du}=e^{u}$ so $\frac{d\left(e^{2A}\right)}{dr}=\frac{du}{dr}\frac{d\left(e^{2A}\right)}{du}=2e^{2A}\frac{dA}{dr}$ . I would never have got that. Thank you.2011-12-05
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    Dumb question: Is this a matrix exponential? Is that why you're worried? Best,2011-12-05
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    Dylan Moreland: I'm just not very good at calculus!2011-12-05
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    Another dumb question then: how come you're calculating Christoffel symbols...? No offence, but that sounds quite advanced if you're having trouble with the chain rule. (P.S. For messages to another user, put "@" before the user name, and they will get notification in their inbox at the upper left of the page.)2011-12-05
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    And by the way: perhaps if you write $u=A(r)$ it's easier to see why $r$ is the variable to differentiate with respect to; $A(r)$ just symbolizes any function of $r$, like $A(r)=r^5$ in the example.2011-12-05
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    Hans Lundmark: Thanks for that. In my own, mathematically inept, way I'm attempting to understand the basics of general relativity. I'm now trying to work through a simple derivation of the Schwarzschild metric. I concur with Einstein when he said, "Do not worry about your difficulties in mathematics. I can assure you mine are still greater." But not as great as mine are!2011-12-06
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    @HansLundmark - I forgot the @. Thanks for your help.2011-12-07
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    Your're welcome, Peter. Good luck with the general relativity! (It's not easy!)2011-12-07

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