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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)

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    @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.2011-07-03

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