User:Corkgkagj/disp

Alternate proof: using properties of limits
The product rule can also be proven using only the properties of limits. The advantage of this method is that it requires no techniques that are "deeper" than the product rule itself. If f and g are both differentiable (and therefore continuous) at x, then
 * $$\frac{\mathrm{d}}{\mathrm{d}x}(fg)=\lim_{h \to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}. \qquad (1)$$

Adding a zero-valued quantity does not effect the equality:
 * $$\frac{\mathrm{d}}{\mathrm{d}x}(fg)=\lim_{h \to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}\,+\,\lim_{h \to 0}\frac{f(x+h)g(x)}{h}(1-1). \qquad (2)$$

The limit law for sums states that
 * $$\lim_{x \to a}\sum_{i=1}^N f_i(x)=\sum_{i=1}^N \lim_{x \to a}f_i(x). \qquad (3)$$

By (3), one may distribute the quantity in (2) as follows:
 * $$\frac{\mathrm{d}}{\mathrm{d}x}(fg)=\lim_{h \to 0}f(x+h)\frac{g(x+h)-g(x)}{h}+\lim_{h \to 0}g(x)\frac{f(x+h)-f(x)}{h}. \qquad (4)$$

The limit law for products states that
 * $$\lim_{x \to a}\prod_{i=1}^N f_i(x)=\prod_{i=1}^N\lim_{x \to a}f_i(x). \qquad (5)$$

By (4) and (5),
 * $$\frac{\mathrm{d}}{\mathrm{d}x}(fg)=\lim_{h \to 0}f(x+h)\cdot\lim_{h \to 0}\frac{g(x+h)-g(x)}{h}\,+\,\lim_{h \to 0}g(x)\cdot\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \qquad (6)$$

Since f is continuous at x,
 * $$\lim_{h \to 0}f(x+h)=f(x). \qquad (7)$$

Since the limit does not effect g(x),
 * $$\lim_{h \to 0}g(x)=g(x). \qquad (8)$$

And the definition of the derivative of a function F with respect to x is
 * $$\lim_{h \to 0}\frac{F(x+h)-F(x)}{h}=\frac{\mathrm{d}F}{\mathrm{d}x}. \qquad (9)$$

One may conclude by (6-9) that
 * $$\frac{\mathrm{d}}{\mathrm{d}x}(fg)=f\frac{\mathrm{d}g}{\mathrm{d}x}+g\frac{\mathrm{d}f}{\mathrm{d}x}$$