# gradient vector of composition of functions

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? $$\nabla{(h\circ f)}(P)=h'(f(P))\nabla f(P)$$

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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 2 '11 at 8:42
Thanks for your big help. Studying math by myself sometimes makes me very confused. :) – CHOI Jul 2 '11 at 9:16
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 3 '11 at 8:50
@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. – Rahul Jul 3 '11 at 8:57
@anon: You should post your comment as an answer. – Rahul Jul 3 '11 at 9:00