If f is differentiable everywhere, then, in terms of f and a, what is the value of: lim x-->a (f(f(x))-f(f(a)))/(x-a)?
That equation simply finds the slope of a secant line between the points (x,f(x)) to (a,f(a)) as x approaches a. When x approaches a, the slope of the secant line begins to approach the slop of f(x) at a. In other words, it finds the slope at a or the derivative at a. Therefore, it equals f'(a).
I'm afraid not. Unless the f(f(x)) and f(f(a)) are typos, the answer is the derivative of f(f(x)) at x = a, NOT the derivative of f(x) at x = a ......
Using the chain rule:
$\displaystyle [f(f(x))]' = f'(f(x)) \times f'(x)$. At x = a, this will equal $\displaystyle f'(f(a)) \times f'(a)$. Making it a little neater, if you knew that f(a) = b, then it will equal $\displaystyle f'(b) \times f'(a)$ ......
This question is a bit more complicated, I think.
Because f is differentiable everywhere it is continuous everywhere.
That means that $\displaystyle \lim _{x \to a} \left( {f(x) - f(a)} \right) = 0$.
Now consider the following:
$\displaystyle \lim _{x \to a} \frac{{f\left( {f(x)} \right) - f\left( {f(a)} \right)}}
{{x - a}} = \lim _{x \to a} \frac{{f\left( {f(x)} \right) - f\left( {f(a)} \right)}}
{{f(x) - f(a)}}\lim _{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}
{{x - a}}$
$\displaystyle \lim _{x \to a} \frac{{f\left( {f(x)} \right) - f\left( {f(a)} \right)}}
{{x - a}} = f'\left( {f(a)} \right)f'(a)$