# multivariable differential for inner product(scalar product)?

In calculus of a single variable, it is well know that the derivative of product $fg$ is $f'g+fg'$. This formula holds also when the function is vector-valued of a single variable, or complex-valued of a complex variable. But if $\mathbf{f,g}$ are general functions from ${\mathbb R}^n$ into ${\mathbb R}^m$, do we have a similar formula for the differential(total derivative) of their inner product(scalar product) $\mathbf{f\cdot g}$ defined as $(\mathbf{f\cdot g)(x)=f(x)\cdot g(x)}$? Thanks.
Maybe we can do as follows. Given a linear transformation $A$ of $\mathbb R^n$ into $\mathbb R^m$ and a vector $\mathbf u$ in $\mathbb R^m$, define $B$ to be a map on $\mathbb R^n$ satisfying $B\mathbf{v=u}\cdot A\mathbf v=A\mathbf v\cdot\mathbf u$ for all $\mathbf v\in\mathbb R^n$. It is clear that $B$ is a linear transformation too; denote it as $\mathbf u\cdot A=A\cdot\mathbf u$. Then if both $\mathbf f$ and $\mathbf g$ are differentiable, it can be proved by definition that $(\mathbf{f\cdot g)'=f'\cdot g+f\cdot g'}$, 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 $A=\mathbf{f'\cdot g+f\cdot g'}$, then $A\mathbf{h=f(x)\cdot g'(x)h+g(x)\cdot f'(x)h}$. So $\mathbf{f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}$ $\bf{=f(x+h)\cdot(g(x+h)-g(x))+g(x)\cdot(f(x+h)-f(x))}$ $=\bf{f(x+h)\cdot(g(x+h)-g(x)-g'(x)h)+g(x)\cdot(f(x+h)-f(x)-f'(x)h)}$ $+\bf{f(x+h)\cdot g'(x)h+g(x)\cdot f'(x)h}$. As a result $0\leq\frac{|\mathbf{f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}-A\mathbf h|}{|\bf h|}$ $=\bf{|f(x+h)\cdot(g(x+h)-g(x)-g'(x)h)+g(x)\cdot(f(x+h)-f(x)-f'(x)h)}$ $+\bf{f(x+h)\cdot g'(x)h+g(x)\cdot f'(x)h-(f(x)\cdot g'(x)h+g(x)\cdot f'(x)h)|/|h|}$ $\leq\bf{|f(x+h)|\frac{|g(x+h)-g(x)-g'(x)h|}{|h|}+|g(x)|\frac{|f(x+h)-f(x)-f'(x)h|}{|h|}}$ $+\bf{\frac{|(f(x+h)-f(x))\cdot g'(x)h|}{|h|}}$. Since $f$ is differentiable at $\bf x$, it is continuous at $\bf x$. Therefore, when $\bf{h\to 0}$, the first term $\to\bf{|f(x)|}\cdot 0=0$ by definition of differential, and the second term $\to 0$ for the same reason. The third term $\leq\bf{\frac{|f(x+h)-f(x)|\|g'(x)\||h|}{|h|}=|f(x+h)-f(x)|\|g'(x)\|}$, so the limit of the this term is 0 when $\bf{h\to 0}$ due to continuity of $\mathbf f$. Then $\lim\limits_{h\to 0}\frac{|\mathbf{f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}-A\mathbf h|}{|\bf h|}=0$ by sequeeze theorem, indicating that $A$ is just the wanted differential.