Need help understanding proof for $g(x) = f(cx) \implies g'(x) = c\cdot f'(cx)$ So I used the definition of a limit on $g(x)$ to get:
$$g'(x) = \lim_{h \to 0} \frac{g(x+h)-g(x)}{h}$$
then subsituted $f(cx)$:
$$g'(x) = \lim_{h \to 0} \frac{f(c(x+h))-f(cx)}{h}$$
and then my textbook says to do this:
$$f(x) = \lim_{h \to 0} \frac{c(f(c(x+h)) - f(cx))}{h}$$
But I don't get how that follows, can anyone help?
 A: It helps to use a substitution. If we let $k = ch$, then as $h \to 0$, we have that $k \to 0$. So:
\begin{align*}
g'(x)
&= \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} \\
&= \lim_{h \to 0} \frac{f(c(x + h)) - f(cx)}{h} \\
&= c\lim_{h \to 0} \frac{f(cx + ch) - f(cx)}{ch} \\
&= c\lim_{k \to 0} \frac{f(cx + k) - f(cx)}{k} \\
&= c f'(cx)
\end{align*}
A: Define $h(x)=cx$ and $g=f\circ h$. Now, $g'(x)=f'(h(x)).h'(x)=f'(cx).c$
A: First off you could use the chain rule stating that $[f(g(x))]'=f'(g(x))g'(x)$ to prove this equality. However using your method, when you substitute $f$, notice how you have $f(c(x+h))-f(cx)$ and when you distribute $c$ you are creating a change in $x$ that is $ch$. Therefore it should be
$$f'=\lim_{h\to0}\frac{f(c(x+h))-f(cx)}{ch}$$
Multiply both sides by $c$ to get (like your textbook says to do)
$$cf'=\lim_{h\to0}\frac{f(c(x+h))-f(cx)}{h}$$
which cancels the $c$ on the denominator of the limit. Then substitute $g$ back in to get
$$cf'=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}=g'$$
A: It ought to say this:
$$
f'(cx) = \lim \limits_{h \to 0} \frac{f(c(x+h)) - f(cx)}{ch}. \tag 1
$$
The point is that
$$
f'(cx) = \lim_{j\to0} \frac{f(cx+j) - f(cx)} j
$$
What this thing is called $j$ that approaches $0$ doesn't matter, so it could be $ch$, where $h$ approaches $0$, and then you have $(1)$.
From $(1)$ we can deduce this:
$$
cf'(cx) = \lim_{h\to0} \frac{f(c(x+h)) - f(cx)} h.
$$
That means $\dfrac d {dx} f(cx) = cf'(cx)$.
