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    Given two functions f and g, let h:=\sqrt{1+f^2e^g}. Calculate nabla h in terms of f, g, nabla f and nabla g.

    What I did was

    \nabla h = (\frac{\partial h}{\partial f}, \frac{\partial h}{\partial g}) = (\frac{e^g f}{\sqrt{1+f^2e^g}} \nabla f, \frac{\frac{1}{2}f^2e^g}{\sqrt{1  +f^2e^g}} \nabla g)

    Figuring that nabla f is the gradient of f, so it's sorta the chain rule. (It is the chain rule?). But evidently I don't really understand what I'm doing, so even if its right, it is bad.
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    Why do you think \nabla h = (\frac{\partial h}{\partial f}, \frac{\partial h}{\partial g})

    Most obviously, nabla h is normally a 3-vector.
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    (Original post by DFranklin)
    Why do you think \nabla h = (\frac{\partial h}{\partial f}, \frac{\partial h}{\partial g})

    Most obviously, nabla h is normally a 3-vector.
    Oh. Hmm. Well I thought h was a function of f and g.
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    And f and g are functions of (x, y, z). Since \nabla h = (\frac{\partial h}{\partial x}, \frac{\partial h}{\partial y}, \frac{\partial h}{\partial z}), the best way to proceed here is to try to find those derivatives. (You will then need to recognize which bits can be rewritten in terms of \nabla f, \nabla g as the end).

    There *are* equivalents of the chain rule, which with practice you can do to solve these problems directly. But it does require some practice and familiarity. (Basically you need to solve problems like this the "long" way until you get to where you can see "oh, \nabla (f^2) is just going to be 2 f \nabla f" etc).
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    (Original post by DFranklin)
    And f and g are functions of (x, y, z). Since \nabla h = (\frac{\partial h}{\partial x}, \frac{\partial h}{\partial y}, \frac{\partial h}{\partial z}), the best way to proceed here is to try to find those derivatives. (You will then need to recognize which bits can be rewritten in terms of \nabla f, \nabla g as the end).

    There *are* equivalents of the chain rule, which with practice you can do to solve these problems directly. But it does require some practice and familiarity. (Basically you need to solve problems like this the "long" way until you get to where you can see "oh, \nabla (f^2) is just going to be 2 f \nabla f" etc).
    Thanks. I'll try that. Although I'm pretty confused over how you knew they were functions of (x, y, z) rather than say (x, y) or (w, x, y z). Is it just convention?
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    Since when do f and g have to be functions of 3 variables?

    http://en.wikipedia.org/wiki/Gradient#Definition
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    It's somewhat convention, but outside of general relativity and the like, I've never seen someone use \nabla and not mean 3D (I'm not sure I've even seen it done in GR, and the Wiki page talks almost exclusively about 3D, although I'm sure I've seen \nabla^2 \phi to refer to \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} - \frac{\partial^2 \phi}{\partial t^2} on occasion).

    Edit: I acknowledge Tom's counter wiki-link, I was referring to the one about the del operator.
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    Okay. So, take

    \frac{\partial h}{\partial x} = \frac{\partial}{\partial x} \sqrt{1+(f(x,y,z))^2e^{(g(x,y,z)  )}}

    =\frac{2 \frac{\partial f}{\partial x} f(x,y,z) e^{(g(x,y,z))} + \frac{\partial g}{\partial x} e^{g((x,y,z))} (f(x,y,z))^2}{\sqrt{1+(f(x,y,z))  ^2e^{g(x,y,z)}}}

    and the others are similar but with ys/zs in the partial derivatives?

    So in the end it's \nabla h=\frac{2 f(x,y,z) e^{(g(x,y,z))} }{\sqrt{1+(f(x,y,z))^2e^{g(x,y,z  )}}} \nabla f + \frac {e^{g((x,y,z))} (f(x,y,z))^2}{\sqrt{1+(f(x,y,z))  ^2e^{g(x,y,z)}}} \nabla g?
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    Looks about right.
 
 
 
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