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mathwonk
#32
May26-05, 03:37 PM
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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....