# differentiable products

• Apr 24th 2007, 12:16 PM
slowcurv99
differentiable products
Suppose that f: I -> R and g: I -> R are differentiable at c element of I and that g(c) is not equal to 0

A) show that (1/g)'= -g'(c)/[g(c)]^2
B) Use part (A) and the product rule to derive the quotient rule
• Apr 24th 2007, 12:49 PM
ThePerfectHacker
Quote:

Originally Posted by slowcurv99
Suppose that f: I -> R and g: I -> R are differentiable at c element of I and that g(c) is not equal to 0

A) show that (1/g)'= -g'(c)/[g(c)]^2

Let g(x) be differenciable at c. And g(c)=0.

Then we can define h(x) = 1/g(x) for x!=c on some open interval containing c. (See proof below).

Proof: Since g(c) is not zero there is an interval containing c so that g(x)!=0 except at c. The reason is because g(x) is differenciable on I so it is continous on I. Thus, |g(x)-g(c)| < e. Thus, g(c) - e< g(x) < g(c) + e. If g(c) >0 then choose e small enough so that g(c) - e>0. And if g(c) < 0 choose e small enough so that g(c) + e<0. On the interval |x-c|< d so c-d<x<c+d. So choose (c-d,c+d) as your interval.
Q.E.D.

This tells us that 1/g(x) is defined on an open interval containing c except possible at c.

We can therefore define h(x) = (1/x) o (g(x)) and use the chain rule.
• Apr 24th 2007, 01:33 PM
Jhevon
Quote:

Originally Posted by slowcurv99
B) Use part (A) and the product rule to derive the quotient rule

(f(x)/g(x))' = (f(x) * 1/g(x))'

...............= f'(x)(1/g(x)) + f(x)(1/g(x))' by the product rule

but from (A), (1/g(x))' = -g'(x)/[g(x)]^2

so (f(x)/g(x))' = f'(x)(1/g(x)) + f(x){-g'(x)/[g(x)]^2}
....................= f'(x)(1/g(x)) – f(x)g'(x)/[g(x)]^2
....................= f'(x)/g(x) - f(x)g'(x)/[g(x)]^2 ..............Now combine the fractions
....................= [f'(x)g(x) – f(x)g'(x)]/g(x)^2 ...............The Quotient Rule