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Let $f$, $g$ and $h$ be any $C^{2}$ scalar functions. Using the standard identities of vector calculus, prove that;

$$ \nabla \cdot \left( f\nabla g \times \nabla h \right) = \nabla f \cdot \left(\nabla g \times \nabla h \right)$$

Here is my working out so far; using identity 8 $$ \nabla \cdot \left( f\nabla g \times \nabla h \right) = \nabla h \cdot \left(\nabla \times f\nabla h \right) - f\nabla g \cdot \left( \nabla \cdot \nabla h \right) $$ and the div of a scalar is a vector hence $$ = \overrightarrow H \cdot \left(\nabla \times f\nabla \overrightarrow G \right) - f\nabla \overrightarrow G \cdot \left( \nabla \cdot \overrightarrow H \right) $$ and then using vector identity 10 gives me $$= \overrightarrow H \cdot \left(f\nabla \times \overrightarrow G +\nabla f \times \overrightarrow G\right) - f\nabla \overrightarrow G \cdot \left( \nabla \cdot \overrightarrow H \right) $$ and now I don't know whats next?

Here are the vector identities listed below

Vector Identities

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    $\begingroup$ @Daryl Hi Daryl, please see my updated post. $\endgroup$ – amanda Jun 1 '13 at 7:45
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HINT: Use identity 7 with $\mathbf{F}=\nabla g\times\nabla h$.

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  • $\begingroup$ The question is a lot easier than i thought, Thankyou $\endgroup$ – amanda Jun 1 '13 at 7:50
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HINT: Use Identity 8 with $\mathbf{F} = f\nabla g$, $\mathbf{G} = \nabla h$: $$ \nabla \cdot (f\nabla g\times \nabla h)= \nabla h\cdot \big(\nabla \times (f\nabla g)\big)- f\nabla g\cdot \big(\nabla\times (\nabla h)\big). $$ Notice $\nabla \times (\nabla h) = 0$. Then use Identity 10 on $\nabla \times (f\nabla g)$.

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  • $\begingroup$ Hi Shuhao, that was exactly what I did before writing this question but I thought I was wrong because I couldn't make out anything of it after that. I have updated my post to show my attempt, and now I am unsure of what is next? $\endgroup$ – amanda Jun 1 '13 at 7:41
  • $\begingroup$ @amanda In the second term it is $\nabla \times (\nabla h)$, not $\nabla \cdot (\nabla h)$. $\endgroup$ – Shuhao Cao Jun 1 '13 at 15:54
  • $\begingroup$ Ow, I see where I went wrong, thankyou for that Shuhao $\endgroup$ – amanda Jun 2 '13 at 15:36

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