proof of product rule of derivatives using first principle? How can I prove the product rule of derivatives using the first principle?
$$\frac{d (f(x) g(x))}{d x}
   = \left( \frac{d f(x)}{d x} g(x) + \frac{d g(x)}{d x} f(x) \right)$$
Sorry if i used the wrong symbol for differential (I used \delta), as I was unable to find the straight "d" on the web.
 A: There are two ways of stating the first principle. 
The first one is $$\frac{{\rm d}f(x)}{{\rm d}x} =\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}.$$
Then \begin{align}
\frac{{\rm d}\big(f(x)g(x)\big)}{{\rm d}x}
&=\lim_{h\to 0} \frac{f(x+h)g(x+h) -f(x)g(x)}{h}\\
&=\lim_{h\to 0} \frac{f(x+h)g(x+h)\color{blue}{-f(x)g(x+h)+f(x)g(x+h)}-f(x)g(x)}{h}\\
&=\lim_{h\to 0} \left( \frac{f(x+h)\color{blue}{-f(x)}}{h} \cdot g(x+h)
 +f(x) \cdot \frac{\color{blue}{g(x+h)} -g(x)}{h} \right)\\
&=\lim_{h\to 0} \frac{f(x+h)\color{blue}{-f(x)}}{h} \cdot \lim_{h\to 0} g(x+h)
 +\lim_{h\to 0} f(x) \cdot \lim_{h\to 0} \frac{\color{blue}{g(x+h)}-g(x)}{h}\\
&=\frac{{\rm d}f(x)}{{\rm d}x} \cdot g(x)
 +f(x) \cdot \frac{{\rm d}g(x)}{{\rm d}x}.\end{align}
The second way is $$f'(a) =\lim_{x\to a} \frac{f(x)-f(a)}{x-a}.$$
Then \begin{align}
(fg)'(a) &=\lim_{x\to a} \frac{f(x)g(x) -f(a)g(a)}{x-a}\\
&=\lim_{x\to a} \frac{f(x)g(x)\color{blue}{-f(a)g(x)+f(a)g(x)}-f(a)g(a)}{x-a}\\
&=\lim_{x\to a} \left( \frac{f(x)\color{blue}{-f(a)}}{x-a} \cdot g(x)
 +f(a) \cdot \frac{\color{blue}{g(x)}-g(a)}{x-a} \right)\\
&=\lim_{x\to a} \frac{f(x)\color{blue}{-f(a)}}{x-a} \cdot \lim_{x\to a} g(x)
 +\lim_{x\to a} f(a) \cdot \lim_{x\to a} \frac{\color{blue}{g(x)}-g(a)}{x-a}\\
&= f'(a) \cdot g(a) +f(a) \cdot g'(a).\end{align}
