If $\frac{d\;f(x)}{dx}=k\;g(x)$, then is $\frac{\left(\frac{d\;f(x)}{k}\right)}{dx}= g(x)$ or $\frac{d\left(\frac{f(x)}{k}\right)}{dx}= g(x)$?

Question

If $$\frac{d\;f(x)}{dx}=k\;g(x)$$ where $f(x)$ is a function of $x$ and $g(x)$ is another function of $x$, then is $$\frac{\left(\dfrac{d\;f(x)}{k}\right)}{dx}= g(x)$$ or $$\frac{d\left(\dfrac{f(x)}{k}\right)}{dx}= g(x)$$true?

Answer 1

Yes it is true. A proof would be as follows.$$\frac{df}{dx}=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=k \,g(x)$$So$$\frac d{dx}\left(\frac{f(x)}k\right)=\lim_{h\to 0} \frac{\frac{f(x+h)}k-\frac{f(x)}k}{h}=\frac1k \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=\frac1kk\,g(x)=g(x)$$ This is why you can take a constant into a derivative (more commonly, a constant can be taken out of a derivative). E.g. something like $\frac {d}{dx}(2x)=2\frac{d}{dx} x$.

Answer 2

$$\frac{d\space f(x)}{d\space x}=k\space g(x) \implies \frac{d(\frac{f(x)}{k})}{d\space x}= \frac {1}{k}\frac{d\space f(x)}{d\space x}=\frac {1}{k}k\space g(x)= g(x)$$