You have two conflicting goals here. If $y$ is arbitrary, then $x^y$ only
makes sense for $x>0$. Imagine, for example, that $y = \frac{1}{2}$.
Then $x^y = \sqrt{x}$ - what does that mean for negative $x$?
Note that switching to complex numbers doesn't help much - negative numbers do have
square roots then, but those are non-unique, and what's worse, the number
of solutions is highly dependent on $y$! E.g., $y^n = x$ has $n$ solutions in
$\mathbb{C}$. Which is $x^\frac{1}{n}$ supposed to be?
So you'll have to distinguish between two cases. One is $f(x)^{g(x)}$ for
positive $f$, and the other is $f(x)^k$ for constants $k \in \mathbb{Z}$
(i.e., no fractional exponents). You could generalize the second case to
$f(x)^{g(x)}$ for functions $g$ which take only integral values, but since
such functions are either constant or non-continuous, that case isn't really
interesting for purposes of differentiation, I think.
BTW, a far more interesting (and maybe solvable!) question is how to deal with
non-negative $f$, which nevertheless may take the value zero. $f(x)^{g(x)}$ is perfectly well-defined for those, but you'll still run into problems with the logarithm. Now, in some cases these problems are due to the fact that the derivative does, in fact, not exists at these points. But not in al cases! For example, $f(x) = x^2$ has derivative $0$ at $x=0$. The reason is, basically, that since $g$ is constant in this case, then $g' ln f$ doesn't matter, because $g' = 0$, and similarly for $f'\frac{g}{f}$. But you can't just cancel things that way in all cases - that will produce wrong answers sometimes, because it actually depends on how fast things go to zero respectively infinity.
You might ask, then, why the non-uniqueness mentioned above doesn't prevent us from sensibly defining $\sqrt[x]{x}$ - after all, $y^n = x$ has two solutions for positive $x$ even in \mathbb{R}$. The reason is twofold
The number of solutions doesn't explode as badly. We have one solution of $y^n = x$ for odd $n$, and two for even $n$.
There's an order on $\mathbb{R}$, which makes the definition of $\sqrt[n]{x}$ as the (unique!) positive solution of $y^n = x$ quite natural.
The effect of (1) and (2) is, for example, that while it's not true that $\sqrt[n]{x^n} = x$, we do get at least that $\sqrt[n]{x^n} = |x|$. Trying to do the same over the complex numbers fails horribly. We could attempt to define $\sqrt[n]{x}$ as the solution of $y^n =x$ with the smallest angle (assuming we agree to measure angles counter-clockwise from the real axis). But then an $n$-th root always has an angle smaller than $\frac{2\pi}{n}$, so $\sqrt[n]{x^n}$ and $x$ would have very little in common except that their $n$-th power is $x^n$.
$f(g(x))$ = $6x$ where $f(x)$ = $2x$ and $g(x)$ = $3x$ then $f'(g(x))$ = 6. Similarly let $3x$ = $u$ in which case $f(g(x))$ = $2(u)$. Following the chain rule: $d/dx$ $f(g(x))$ = $f´(g(x))$ * $g´(x)$ where $f'(g(x))$ = 2 and $g'(x)$ = $u'$ = 3. Thus 2*3 = 6.
Maybe a problem you're encountering is that above $f(x)$ = $x^2$ and $g(x)$ = $x^3$. So $f'(x) = 2x$ and $g'(x)$ = $3x^2$ But $f(g(x))$ = $x^6$ and $f'(g(x)$ = $6x^5$. Similarly $f'(g(x))$ = $2(x^3)*3*x^2$ = 6x^5.
I hope I understood what you were confused about. Hopefully that helped.
Best Answer
Not really. Actually, what you want is uniform convergence and majorant series.
An important property is a series might have is being majorant.
Yet another important case scenario is uniform convergence:
We usually say $N$ is independent of the choice of $x$, too. You can picture this behaviour as follows: Each partial sum is always contained in the strip inside $f(x)+\epsilon$ and $f(x)-\epsilon$ of width $2\epsilon$.
In particular, every majorant series converges uniformly. This is known as Weierstrass' $M$ criterion. For majorant series, the following is valid:
This stems from
You can read this in much more detail, and find proofs, in (IIRC) Apostol's Calculus (Vol.1)