Derivative+Rules

Differentiation - the Product Rule
p'(x) = f'(x)g(x) + f(x)g'(x)
 * Product Rule:** If p(x) = f(x)g(x) then

h || = |||| __f(x+h)g(x+h) -f(x)g(x)__ h ||  |||| || h || h || + |||| __f(x)g(x+h) - f(x)g(x)__ h || h || + |||| __f(x)[g(x+h) - g(x)]__ h || h || g(x+h) || + || f(x) || __[g(x+h) - g(x)]__ h ||  Now taking the limit as h approaches zero implies the stated rule: p'(x) = f'(x)g(x) + f(x)g'(x)
 * Proof:** Observe
 * __p(x+h)-p(x)__
 * || = |||||||||| __f(x+h)g(x+h) - f(x)g(x+h) + f(x)g(x+h) - f(x)g(x)__
 * || = |||| __f(x+h)g(x+h) - f(x)g(x+h)__
 * || = |||| __[f(x+h) - f(x)] g(x+h)__
 * || = || __f(x+h) - f(x)__

Differentiation - The Quotient Rule
 dx || || f g ||  || (x) || = ||   || lim h0 |||||| __f(x+h)/g(x+h) - f(x)/g(x)__ h || || h0 || f(x+h)g(x) - f(x)g(x+h) g(x+h)g(x)h ||  || (subtraction of fractions) || h0 || f(x+h)g(x) - f(x)g(x) + f(x)g(x) - f(x)g(x+h) g(x+h)g(x)h ||  || (we added & subtracted f(x)g(x)) || h0 || [f(x+h) - f(x)]g(x) - f(x)[(g(x+h) - g(x)] g(x+h)g(x)h ||  || (a little algebra) || If we recognize the difference quotients for f and g in this last expression, we see that taking the limit as h0 replaces them by the dreivatives f ' (x) and g ' (x). Further, since g is differentiable, it is also continuous, and so g(x+h)g(x) as h0. Putting this all together gives dx ||  || f g ||  || (x) || = || f ' (x) g(x) - f(x) g ' (x) [g(x)]^2 ||   ||, || which is the quotient rule. 
 * Proof** By the definition of the derivative,
 * d
 * || = ||  || lim
 * || = ||  || lim
 * || = ||  || lim
 * ||  ||   ||   ||   ||   ||   ||   ||   ||  ||
 * d