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 Sci Advisor HW Helper P: 9,453 you want to know the proof of the chain rule/ look: definition of the derivative of f at a, is it is a linear function L such that L(v) is tangent to f(a+v)-f(a) at v= 0. i.e. the difference quotient [f(a+v) - f(a) - L(v)]/|v| approaches zero as v does. call a function o(v) such that o(v)/|v| goes to zero as v does "little oh", and write it o(v). A function such that the quotient O(v)/|v| is bounded as v approaches zero, "big oh" and write it as O(v). Then basic ruelks are these: linear combinations of O's are also O, and also for o's, and compositions of o's and O's are always o if even one "factor" is o. and a product of two O's is o. Then the chain rule is as follows; assume L is the derivative of f and M is the derivative of g, then f(a+v)) - f(a) - L(v) = o(v), so f(a+v) - f(a) = L(v) + o(v). Hence M(f(a+v) - f(a)) =M(L(v)) + M(o(v)). since f(a+v) = f(a) + [f(a+v)-f(a)], hence we have g(f(a+v)) - g(f(a)) -M([f(a+v)-f(a)]) = o(f(a+v)- f(a)) = o(O(v)) = o(v). bu also M(f(a+v) - f(a)) =M(L(v)) + M(o(v)), from above, so g(f(a+v)) - g(f(a)) -M([f(a+v)-f(a)]) = g(f(a+v)) - g(f(a)) - M(L(v)) + M(o(v)) = o(v). hence g(f(a+v)) - g(f(a)) - M(L(v)) = -M(o(v)) + o(v) = o(v) + o(v) = o(v). hence by definition, the derivative of g(f) at a is M(L). i.e. the derivative of a composition is the composition, as linear maps, of the derivatives. hence as matrices it is dot product as you are computing above: ie. dw/ds = (dw/dx,dw/dy/dw/dz).(dx/dt, dy/dt, dz/dt), and so on....