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Let $U \subset \mathbb{R}^n$ be open and let $f:U \to \mathbb{R}$ and $h:\mathbb{R}\to \mathbb{R}$ be differentiable functions.

How can I prove the following equation? $$\nabla{(h\circ f)}(P)=h'(f(P))\nabla f(P)$$

t.b.
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CHOI
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    You can prove it one component at a time using the one-dimensional chain rule. Hold all of the components of $ x $ fixed except for $ x_i $; then $ \partial (h \circ f) / \partial x_i = (h' \circ f) , \partial f / \partial x_i $. – anon Jul 02 '11 at 08:42
  • Thanks for your big help. Studying math by myself sometimes makes me very confused. :) – CHOI Jul 02 '11 at 09:16
  • Out of curiosity: Why do people insist on writing the non-suggestive $\nabla$ symbol for gradient? The OP wrote grad, which is perfectly fine as well, in my opinion. I rejected one edit suggesting that same change because it is not clear to me that people having trouble with this question are familiar with the $\nabla$-notation. Now I see that someone else suggested on the same change and it was approved. I'm pretty sure that I learned about $\nabla$ quite a bit after I learned about divergence, rotation and gradient. – t.b. Jul 03 '11 at 08:50
  • @Theo: At least in certain physics and engineering circles, it's far more common to denote gradient by the symbol $\nabla$ than by $\operatorname{grad}$. The suggestiveness of a symbol is surely a function of one's familiarity with it. –  Jul 03 '11 at 08:57
  • @anon: You should post your comment as an answer. –  Jul 03 '11 at 09:00
  • @Rahul: I don't think so: $\nabla$ is a symbol that could mean anything if one doesn't know it while if one knows the word gradient, one cannot do otherwise than read "grad" as gradient. I'm aware that $\nabla$ is more common, but still I don't see the advantage of replacing grad by it in the present case. – t.b. Jul 03 '11 at 09:08

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