The Chain Rule, death to anyone that breaks the rule

[mathwonk]

the graph is tangent to the x axis.

as to your second question, do you know about limits?

you seem to need some review of this basic topic. the fundamental idea is that a limit has a priori nothing at all to do with the value at the point. so the fact that (f(x)-f(a))/(x-a) is undefiend at x= a has nothing at all to do with the limit at x=a.

for example, (x-a)/(x-a) has a limit as x goes to a, even though the bottom goes to zero.

 

[Cyrus]

Right, that's why I asked earlier about having to use L'hospitals rule for the limit to be equal to a defined value as the denominator and numerator go to zero. Oh, opps, READ THE FINE PRINT! IT says in my clac 1 book as x approaches a but NOT EQUAL TO A! Ok, I get it! Ill just blame the tv with the miss universe pagent in the background for the sudden lapse of judgement! Aye caramba, I picked the wrong major!

 

[mathwonk]

it makes no sense to speak of using l'hopitals rule to evaluate derivatives since you must know what a derivative is before even stating l'hopitals rule.

you may need to go back pretty much to the beginning, and learn about limits and derivatives properly. i recommend spivak's calculus book.

 

[Cyrus]

Its been a while since I deat with limits. But I take back what I said about l'hopitals rule. The limit is as x approaches a, but does not EQUAL it. I forgot that minor, but important detail. Hopefully, were both on the same page on that issue now though.

 

[Cyrus]

See here's what I mean though,

If we have as you state: [f(x)-f(a) - f'(a)(x-a)]/(x-a)] . Then we can call this the same as, [f(x)-f(a) - f'(a)(x-a)]= F(x) and (x-a) as G(x). Then we can re-write this limit as F(X)/G(X) as X approaches a. The consequence is that it converges to zero over zero, and thus my earlier statement about l'hospitals rule. And the derivative of the bottom function would be d/dx (x -a) , which is just one, and the numerator would be f'(x)-f'(a) - f'(a)(x'), which is just f'(a) - 0 - f'(a) = 0.

Thus, you get zero over 1, and the limit converges to zero. I am a little rusty, I may be wrong though. My whole issue is that you are talking about a limit that approaches zero in the denominator and numerator, and that's a problem.

 

[Cyrus]

I have to go to bed now, but I will put up a simple example to better show my question. For instance, the definition of a derivative says the limit as x-->0 of
[f(a+x)-f(a)]/(x-a)

If we have something like, f(x)= x^2, and a=1, then the limit simplifies and you can factor the x-a term out of the denominator, thus elminating the divison by zero problem. But this works only for nice functions like this, and does not work for all functions in general, i.e. its not always possible to elmininate the division by zero.