# 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.
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.
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.