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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?

Michael Hardy
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Anmol Bhullar
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    The last line show be $\color{red}{g'}(x) = \lim \limits_{h \to 0} \dfrac{c(f(c(x+h)) - f(cx))}{\color{red}{c}h}$. It just means that the author is multiplying $c$ to both numerator, and denominator. – user49685 Mar 25 '16 at 16:31
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    Something the author might have left out as well is the assumption that $f$ is differentiable (i.e. $f$ has a derivative). – parsiad Mar 25 '16 at 16:35

4 Answers4

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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*}

Adriano
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Define $h(x)=cx$ and $g=f\circ h$. Now, $g'(x)=f'(h(x)).h'(x)=f'(cx).c$

Martín Vacas Vignolo
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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 $$cf'=\lim_{h\to0}\frac{g(x+h)-g(x)}{h}=g'$$

Will Fisher
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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 $\dfrac d {dx} f(cx) = cf'(cx)$.

Michael Hardy
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