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Webif h(x) = f [g(x)], then prove that ∇h(a) = ∑k=1n Dkf (b) ∇gk(a) You can't do h′(a) = ∇h(a)∘a because h is a scalar and a is a vector. Write h(x) as h(x) = f (g1(x),g2(x),...,gn(x)) Then ∇h = (∂x1∂h,..., ∂xn∂h) ... If h(x) = f (g(f (x))) is bijective, what do we know about f,g? Your proof is fine. It's also worth noting ... WebFeb 23, 2015 · Initial note: I can see you are using the product rule when only three functions are involved, but why stop there? Why not generalize it? Generalizing it will solve your specific problem and also when you would have $4, 5, \ldots, n$ functions. economy of burma improvement
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