# Thread: multivariable differential for inner product(scalar product)?

1. ## multivariable differential for inner product(scalar product)?

In calculus of a single variable, it is well know that the derivative of product $\displaystyle fg$ is $\displaystyle f'g+fg'$. This formula holds also when the function is vector-valued of a single variable, or complex-valued of a complex variable. But if $\displaystyle \mathbf{f,g}$ are general functions from $\displaystyle {\mathbb R}^n$ into $\displaystyle {\mathbb R}^m$, do we have a similar formula for the differential(total derivative) of their inner product(scalar product) $\displaystyle \mathbf{f\cdot g}$ defined as $\displaystyle (\mathbf{f\cdot g)(x)=f(x)\cdot g(x)}$? Thanks.

2. Maybe we can do as follows. Given a linear transformation $\displaystyle A$ of $\displaystyle \mathbb R^n$ into $\displaystyle \mathbb R^m$ and a vector $\displaystyle \mathbf u$ in $\displaystyle \mathbb R^m$, define $\displaystyle B$ to be a map on $\displaystyle \mathbb R^n$ satisfying $\displaystyle B\mathbf{v=u}\cdot A\mathbf v=A\mathbf v\cdot\mathbf u$ for all $\displaystyle \mathbf v\in\mathbb R^n$. It is clear that $\displaystyle B$ is a linear transformation too; denote it as $\displaystyle \mathbf u\cdot A=A\cdot\mathbf u$. Then if both $\displaystyle \mathbf f$ and $\displaystyle \mathbf g$ are differentiable, it can be proved by definition that $\displaystyle (\mathbf{f\cdot g)'=f'\cdot g+f\cdot g'}$, a very familiar form. But I'm not sure if it is correct, or there is any other better results. Any comment or suggestion is welcomed.

3. I finally found a related discussion in Charles Chapman Pugh's "Real Mathematical Analysis", P274. I'm glad my conclusion is the same as the one in this book. But my proof seems somewhat different, as follows: Denote $\displaystyle A=\mathbf{f'\cdot g+f\cdot g'}$, then $\displaystyle A\mathbf{h=f(x)\cdot g'(x)h+g(x)\cdot f'(x)h}$. So $\displaystyle \mathbf{f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}$$\displaystyle \bf{=f(x+h)\cdot(g(x+h)-g(x))+g(x)\cdot(f(x+h)-f(x))}$$\displaystyle =\bf{f(x+h)\cdot(g(x+h)-g(x)-g'(x)h)+g(x)\cdot(f(x+h)-f(x)-f'(x)h)}$$\displaystyle +\bf{f(x+h)\cdot g'(x)h+g(x)\cdot f'(x)h}. As a result \displaystyle 0\leq\frac{|\mathbf{f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}-A\mathbf h|}{|\bf h|}$$\displaystyle =\bf{|f(x+h)\cdot(g(x+h)-g(x)-g'(x)h)+g(x)\cdot(f(x+h)-f(x)-f'(x)h)}$$\displaystyle +\bf{f(x+h)\cdot g'(x)h+g(x)\cdot f'(x)h-(f(x)\cdot g'(x)h+g(x)\cdot f'(x)h)|/|h|}$$\displaystyle \leq\bf{|f(x+h)|\frac{|g(x+h)-g(x)-g'(x)h|}{|h|}+|g(x)|\frac{|f(x+h)-f(x)-f'(x)h|}{|h|}}$$\displaystyle +\bf{\frac{|(f(x+h)-f(x))\cdot g'(x)h|}{|h|}}$. Since $\displaystyle f$ is differentiable at $\displaystyle \bf x$, it is continuous at $\displaystyle \bf x$. Therefore, when $\displaystyle \bf{h\to 0}$, the first term $\displaystyle \to\bf{|f(x)|}\cdot 0=0$ by definition of differential, and the second term $\displaystyle \to 0$ for the same reason. The third term $\displaystyle \leq\bf{\frac{|f(x+h)-f(x)|\|g'(x)\||h|}{|h|}=|f(x+h)-f(x)|\|g'(x)\|}$, so the limit of the this term is 0 when $\displaystyle \bf{h\to 0}$ due to continuity of $\displaystyle \mathbf f$. Then $\displaystyle \lim\limits_{h\to 0}\frac{|\mathbf{f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}-A\mathbf h|}{|\bf h|}=0$ by sequeeze theorem, indicating that $\displaystyle A$ is just the wanted differential.

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# total derivative of inner product

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