Suppose that f and g are r^th -order differentiable and that the composite h = g

ID: 3109497 • Letter: S

Question

Suppose that f and g are r^th -order differentiable and that the composite h = g compositefunction f makes sense. A partition divides a set into nonempty disjoint subsets. Prove the Higher Order Chain Rule, (D^r h)_p = sigma^r _k=1 sigma_mu element (k, r) (D^k g)_q compositefunction (D^mu f)_p where mu partitions {1, ..., r} into k subsets, and q = f(p). In terms of r-linear transformations, this notation means (D^r h)_p (v1, ..., v_r) = sigma^r _k=1 sigma _mu (D^k g)_q ((D^|mu_1| f)_p (v mu_1), ..., (D^|mu_k| f)_p (v mu_k)) where |mu_i| = notequalto mu_i and v_mu_i and the order in which the partition blocks mu_1, ..., mu_k occur are irrelevant.)

Explanation / Answer

Theorem (Chain Rule): If f:RnRm is differentiable at a, and g:RmRp is differentiable at f(a), then the composition gf:RnRp is differentiable at a, and D(gf)(a)=Dg(f(a))Df(a).

Proof: Suppose f:RnRm

is differentiable at a, and g:RmRp is differentiable at f(a).

This means that

(1) limh0|f(a+h)f(a)Df(a)(h)| / |h|=0,

and that

(2) limh0|g(f(a)+h)g(f(a))[Dg(f(a))](h)| / |h|=0.

Note then that

|(gf)(a+h)(gf)(a)[Dg(f(a))Df(a)](h)| /|h|=

|g(f(a+h))g(f(a))Dg(f(a))[Df(a)(h)]| /|h|=

|g(f(a+h))g(f(a))Dg(f(a))[f(a+h)f(a)f(a+h)+f(a)+Df(a)(h)]| /|h|

|g(f(a+h))g(f(a))Dg(f(a))[f(a+h)f(a)]|/|h|+|Dg(f(a))[f(a+h)f(a)Df(a)(h)]|/|h|.

But by (2), for every >0

|g(f(a+h))g(f(a))Dg(f(a))[f(a+h)f(a)]| /|h|<

|f(a+h)f(a)||h|= |f(a+h)f(a)Df(a)(h)+Df(a)(h)| / |h|

|f(a+h)f(a)Df(a)(h)|/|h|+|Df(a)(h)| /|h|.Now, since Df(a)

and Dg(f(a)) are linear transformations, we also have that for certain M,NR,

|Df(a)(h)|<M|h|

and

|Dg(f(a))[f(a+h)f(a)Df(a)(h)]|<N|f(a+h)f(a)Df(a)(h)|.

This in turn gives first that

|Df(a)(h)| /|h|<M,

and also, by using (1), that for every >0 we can find >0 such that |h|< implies that |f(a+h)f(a)Df(a)(h)|<|h|, from which we also obtain

|f(a+h)f(a)Df(a)(h)| /|h|< and

|Dg(f(a))[f(a+h)f(a)Df(a)(h)]|/|h|< N|f(a+h)f(a)Df(a)(h) /||h|< N|h|/ |h|=N.

Finally, piecing all of the above together (from the last three inequalities we can conclude that both terms on the right hand side of the original inequality are arbitrarily small) gives the conclusion:

limh0|(gf)(a+h)(gf)(a)[Dg(f(a))Df(a)](h)| /|h|=0.

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