# Real Analysis Differential Proof

• September 23rd 2008, 03:21 PM
Caity
Real Analysis Differential Proof
I seem to be having a problem understanding delta epsilon proofs so any explanations would be great if you can do this problem. Its been about 12 years since I took Calculus so I'm having to go back and review alot and proofs have always been a problem for me. I started it using the definition and h->0 but really I'm totally confused with this problem.

Suppose the f is differentiable at a and f'(a) does not = 0.
a) Using the theorem lim x->a f(x)/g(x) = L/M and the definition of the derivative, prove directly (1/f)'(a) = -f'(a)/f^2(a).
b) Use part a) and the product rule to prove the quotient rule.
• September 23rd 2008, 06:31 PM
Jhevon
Quote:

Originally Posted by Caity
I seem to be having a problem understanding delta epsilon proofs so any explanations would be great if you can do this problem. Its been about 12 years since I took Calculus so I'm having to go back and review alot and proofs have always been a problem for me. I started it using the definition and h->0 but really I'm totally confused with this problem.

Suppose the f is differentiable at a and f'(a) does not = 0.
a) Using the theorem lim x->a f(x)/g(x) = L/M and the definition of the derivative, prove directly (1/f)'(a) = -f'(a)/f^2(a).

remember the definition of the derivative of a function (if it exists):

$f'(a) = \lim_{x \to a} \frac {f(x) - f(a)}{x - a}$

or alternatively,

$f'(x) = \lim_{h \to 0} \frac {f(x + h) - f(x)}h$

use $\frac 1{f(x)}$ as your function in either of the expressions above to find the derivative by defintion

Quote:

b) Use part a) and the product rule to prove the quotient rule.
the quotient rule gives the derivative of rational functions, that is, functions of the form $\frac {f(x)}{g(x)}$, where $f(x), g(x) \ne 0$ are functions

note that this can be written as $f(x) \cdot \frac 1{g(x)}$

so you can use the product rule to find the derivative of this product. note that you found the formula to differentiate things like $\frac 1{g(x)}$ in part (a)
• September 24th 2008, 07:17 AM
Caity
Thanks for the help Jhevon. I tried what you suggested but somehow I keep getting zero in the denominator. For part a) I got

lim f(a+h) - f(a) / f(a+h)f(a)h = 0/0
h-> 0

on part b) I ended up with

lim g(x+h)f(x+h) - f(x)g(x) / g(x+h)g(x)h
h-> 0

that one doesnt make sense to me at all...
• September 29th 2008, 10:44 AM
Jhevon
Quote:

Originally Posted by Caity
Thanks for the help Jhevon. I tried what you suggested but somehow I keep getting zero in the denominator. For part a) I got

lim f(a+h) - f(a) / f(a+h)f(a)h = 0/0
h-> 0

you can't just take the limit like that, Caity, you have to simplify first.

note that this is $\lim_{h \to 0} \frac {\frac {f(a) - f(a + h)}h}{f(a + h)f(a)} = \lim_{h \to 0} \frac {- \frac {f(a + h) - f(a)}h}{f(a + h)f(a)}$

anything look familiar in that? now take the limit

Quote:

on part b) I ended up with

lim g(x+h)f(x+h) - f(x)g(x) / g(x+h)g(x)h
h-> 0

that one doesnt make sense to me at all...
now that you (hopefully) know how to do (a), try this again. by the way, they said use the product rule, not the defintion