if 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
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Math is a way of solving problems by using numbers and equations.
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Integration, differentiation) have to be included. Greatest calculator ever created. This is a lovely application for people who are not very good at maths and finds difficulty and some problems in solving hard as well as tricky questions of Maths.