Differentiable function satisfying $f(x+a) = bf(x)$ for all $x$ This is an exercise from Apostol Calculus, (Exercise 10 on page 269).  

What can you conclude about a function which has derivative everywhere and satisfies an equation of the form
  $$ f(x+a) = bf(x) $$
  for all $x$, where $a$ and $b$ are positive constants?

The answer in the back of the book suggests that we should conclude $f(x) = b^{x/a} g(x)$ where $g(x)$ is a periodic function with period $a$.  I'm not sure how to arrive at this.
One initial step is to say, by induction,
$$ f(x+a) = bf(x) \implies f(x+na) = b^n f(x)$$
for all $x$.  I'm not sure what to do with this though.  I'm also not clear how to use the differentiability of $f$.  (If I write down the limit definition of the derivative then I end up with a term $f(x+h)$, but I cannot use the functional equation on that since the functional equation is for a fixed constant $a$.)
 A: One trivial solution that doesn't use the differentiability of $ f(x) $:
From $ f(x+na)=b^nf(x) $, letting $ y=x+na $, and requiring that $ x \in [0,a) $ and $ n = \left\lfloor \frac{y}{a} \right\rfloor $ we get the following equivalent definition of $ f $:
$$ f(y)=b^{\frac{y-\left(y-\left\lfloor \frac{y}{a} \right\rfloor a\right)}{a}}f\left (y- \left\lfloor \frac{y}{a} \right\rfloor a \right) $$
By letting $ g(y)=b^{-\frac{\left( y-\left\lfloor \frac{y}{a} \right\rfloor a\right)}{a}}f\left(y-\left\lfloor \frac{y}{a} \right\rfloor a \right) $ noting that $ g $ is periodic with a period of $ a $:
$$ f(y)=b^{\frac{y}{a}}g(y) $$
A: $$f(x+a)=bf(x)$$
$$f'(x+a)=bf'(x)$$
Then
$$\frac{f'(x+a)}{f(x+a)}=\frac{f'(x)}{f(x)}$$
Let
$$\frac{f'(x)}{f(x)}=h(x)$$
$$f(x)=\exp\left(\int{h(x)}{dx}\right)$$
And we require :
$$h(x+a):=h(x)$$
Now from $f(x+a)=bf(x)$ :
$$\exp\left(\int^{x+a}{h(x)}{dx}\right)=b \,\, \exp\left(\int^x{h(x)}{dx}\right)$$
$$\exp\left(\int^{x+a}_x{h(x)}{dx}\right)=b$$
Now your book suggests $f(x)=b^\frac{x}{a}g(x)=e^{\frac{x \ln b}{a}}g(x)$
This gets :
$$f'(x)=\frac{\ln b}{a}e^{\frac{x \ln b}{a}}g(x)+e^{\frac{x \ln b}{a}}g'(x)$$
$$h(x)=\frac{f'(x)}{f(x)}=\frac{\ln b}{a}+\frac{g'(x)}{g(x)}$$
As $g(x)$ and $g'(x)$ are both periodic we can see that this makes $h(x)$ periodic as well.  So the form we derived earlier is simply related to the form the book describes.
And equating $\frac{f'(x+a)}{f(x+a)}=\frac{f'(x)}{f(x)}$ we get
$$\frac{g'(x+a)}{g(x+a)}=\frac{g'(x)}{g(x)}$$
Which is correct if $g(x)$ has a period $a$.
A: The function $g$ defined by $g(x9=b^{-x/a}f(x)$ clearly satisfies $g(x+a)=g(x)$, hence the periodicity. The given differentiality is not used, except that you may use it to conclude that $g$ is differentiable.
