General question about derivatives. Consider a differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$. The derivative of the function at any point can be written as:
\begin{align*}
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\end{align*}
Suppose we have a constant $c > 0$, is it true that:
\begin{align*}
f'(x) = \lim_{h \to 0} \frac{f(x+\frac{h}{c}) - f(x)}{h} \hspace{3ex}?
\end{align*}
Since when dividing $h$ by some constant in the numerator, it will still become arbitrarily small. Or does it follow that:
\begin{align*}
\lim_{h \to 0} \frac{f(x+\frac{h}{c}) - f(x)}{h} = \frac{1}{c}\lim_{h \to 0} \frac{f(x+\frac{h}{c}) - f(x)}{\frac{h}{c}} = \frac{f'(x)}{c} \hspace{3ex} ? 
\end{align*}
I think that the second case is correct, but I still wanted to be 100% sure.
 A: The last equation you wrote is the correct one. You can check with examples. If $f$ is the identity, namely $f(x)=x$. Then 
$$
\lim_{h\rightarrow 0} \frac{f(x+\frac{h}{c})-f(x)}{h}= \lim_{h\rightarrow 0} \frac{x+\frac{h}{c}-x}{h}=\lim_{h\rightarrow 0} \frac{\frac{h}{c}}{h}=\frac{1}{c}.
$$
If $f(x)=x^2$ then 
$$
\lim_{h\rightarrow 0} \frac{f(x+\frac{h}{c})-f(x)}{h}= \lim_{h\rightarrow 0} \frac{x^2+2x\frac{h}{c}+\frac{h^2}{c^2}-x^2}{h}=\lim_{h\rightarrow 0} \frac{2x\frac{h}{c}+\frac{h^2}{c^2}}{h}=\lim_{h\rightarrow 0} \frac{2x}{c}+\frac{h}{c^2}=\frac{2x}{c}.
$$
A: Here is the proof in case you're interested.  It isn't saying much new compared to Anguepa's examples.
$$
\begin{array}{}
&&\displaystyle \lim_{h \to 0} { f(x + h/c) - f(x) \over h} \\
&=& \displaystyle \lim_{h \to 0} \frac1c { f(x + h/c) - f(x) \over h/c} & \text{...basic algebra}\\
&=& \displaystyle \frac1c \lim_{h \to 0}  { f(x + h/c) - f(x) \over h/c} & \text{...pull out the $1/c$ constant}\\
&=& \displaystyle \frac1c \lim_{cg \to 0}  { f(x + g) - f(x) \over g} & \text{...rename $g = h/c$}\\
&=& \displaystyle \frac1c \lim_{g \to 0}  { f(x + g) - f(x) \over g} & \text{...basic property of limit, see below}\\
&=& \displaystyle \frac1c f'(x) &\text{...by definition of $f'(x)$}
\end{array}
$$
The only step that actually involves calculus is the $4$th equal sign, and further explanation will need to invoke the $(\epsilon, \delta)$-based definition of what a limit is.  It's a bit tedious, so hopefully you are OK with leaving it as "basic property of limit."  :)
A: You have to replace $h$ everywhere by $h/c$ for the limit to still be $f'(x).$ Otherwise, it is generally different.
