Abstract Nonsense

Crushing one theorem at a time

Further Properties of the Total Derivative (Pt. II)


Point of Post: This is a continuation of this post.

\text{ }

We now prove a similar theorem for products of real-valued functions

\text{ }

Theorem: Let f_1,\cdots,f_k:U\to\mathbb{R} with U\subseteq\mathbb{R}^n be differentiable at a\in U. Then, f_1\cdots f_k:U\to\mathbb{R} given by (f_1\cdots f_k)(x)=f_1(x)\cdots f_k(x) is differentiable at a and 

\text{ }

\displaystyle D_{f_1\cdots f_k}(a)=\sum_{j=1}^{k}D_{f_1}(a)\cdots D_{f_j}(a)\cdots D_{f_k}(a)

\text{ }

Proof: We play a similar game to the sum proof. Namely, consider K:\mathbb{R}^n\to\mathbb{R}^k given by x\mapsto (f_1(x),\cdots,f_k(x)) and M:\mathbb{R}^k\to\mathbb{R} given by (x_1,\cdots,x_k)\mapsto x_1\cdots x_k and note that f_1\cdots f_k=M\circ K. Now, K is differentiable at a since each coordinate function is differentiable at a and M is differentiable at a since it’s mutlilinear. Thus, by the chain rule we have that f_1\cdots f_k is differentiable at a and D_{f_1\cdots f_k}(a)=D_{M}(K(a))\circ D_K(a). But, using the same logic as for the previous theorem we have that

\text{ }

D_K(a)=\left(\begin{array}{ccc} & \text{Jac}_{f_1}(a)\\ \hline & \vdots & \\ \hline & \text{Jac}_{f_k}(a)\end{array}\right)

\text{ }

And from our corollary concerning the total derivative of multilinear functions we know that

\text{ }

\displaystyle D_K(K(a))(x)=\sum_{j=1}^{k}f_1(a)\cdots x_j\cdots f_n(a)

\text{ }

where evidently x=(x_1,\cdots,x_n). Thus, with this in mind we see that for every x\in\mathbb{R} we have

\text{ }

\displaystyle \begin{aligned}D_{f_1\cdots f_k}(a)(x) &=\left(D_M(K(a))\right)\left(D_K(a)(x)\right)\left(D_M(K(a))\right)\left(\left(\text{Jac}_{f_1}(a)(x),\cdots,\text{Jac}_{f_k}(a)(x)\right)\right)\\ &=\sum_{j=1}^{k}f_1(a)\cdots\text{Jac}_{f_j}(a)\cdots f_{k}(a)\\ &= \sum_{j=1}^{k}f_1(a)\cdots D_{f_j}(a)(x)\cdots f_k(a)\end{aligned}

\text{ }

\text{ }

We now compute the derivative of the inner product of two differentiable functions. In particular:

\text{ }

Theorem: Let f,g:U\to\mathbb{R}^m where U\subseteq\mathbb{R}^n be differentiable at a\in U. Then, the function F:U\to\mathbb{R}  given by F:x\mapsto \left\langle f(x),g(x)\right\rangle (where \left\langle\cdot,\cdot\right\rangle is the usual inner product (i.e. dot product) on \mathbb{R}^m) is differentiable at a and D_F(a)(x)=\left\langle D_f(a)(x),g(a)\right\rangle+\left\langle f(a),D_g(a)(x)\right\rangle

Proof: We play the same old game. Namely define K:\mathbb{R}^n\to\mathbb{R}^{2m} given by x\mapsto (f_1(x),\cdots,f_m(x),g_1(x),\cdots,g_m(x)) where f_j,g_j are the coordinate functions of f,g respectively. We know since each coordinate function is differentiable at a so is K and moreover

\text{ }

D_K(a)=\left(\begin{array}{ccc} & \text{Jac}_{f_1}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{f_m}(a)\\ \hline & \text{Jac}_{g_1}(a) & \\ \hline & \vdots & \\ \hline & \text{Jac}_{g_m}(a) & \end{array}\right)

\text{ }

We also know from a past corollary that \langle\cdot,\cdot\rangle:\mathbb{R}^{2m}\to\mathbb{R} is differentiable everywhere and

\text{ }

\displaystyle D_{\langle\cdot,\cdot\rangle}((x,y))(z)=\left\langle x,z\right\rangle+\left\langle z,y\right\rangle

\text{ }

(where of course we use shorthand notation meaning x,y\in\mathbb{R}^m). Noticing then that \langle\cdot\cdot\rangle\circ K=F we may conclude from the chain rule that F is differentiable at a and D_F(a)=D_{\langle\cdot,\cdot\rangle}(K(a))\circ D_K(a). Thus,

\text{ }

\begin{aligned}D_F(a)(x)&=\left(D_{\langle\cdot,\cdot\rangle}(K(a))\right)\left(D_K(a)\right)\\ &=\left(D_{\langle\cdot\cdot\rangle}(K(a))\right)\left(\text{Jac}_{f_1}(a)(x),\cdots,\text{Jac}_{f_m}(a)(x),\text{Jac}_{g_1}(a)(x),\cdots,\text{Jac}_{g_m}(a)(x)\right)\\ &=\left\langle \left(\text{Jac}_{f_1}(a)(x),\cdots,\text{Jac}_{f_m}(a)(x)\right),\left(\text{Jac}_{g_1}(a)(x),\cdots,\text{Jac}_{g_m}(a)(x)\right)\right\rangle\\ &= \left\langle \text{Jac}_{f}(a)(x),g(a)\right\rangle+\left\langle f(a),\text{Jac}_{g}(a)(x)\right\rangle\\ &= \left\langle D_f(a)(x),g(a)\right\rangle+\left\langle f(a),D_g(a)(x)\right\rangle\end{aligned}

\text{ }

since x was arbitrary the conclusion follows. \blacksquare

\text{ }

\text{ }

References:

1. Spivak, Michael. Calculus on Manifolds; a Modern Approach to Classical Theorems of Advanced Calculus. New York: W.A. Benjamin, 1965. Print.

Advertisements

May 26, 2011 - Posted by | Analysis | , , , , ,

1 Comment »

  1. […] Point of Post: This is a continuation of this post. […]

    Pingback by Further Properties of the Total Derivative (Pt. II) « Abstract Nonsense | May 26, 2011 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: