Is this observation correct?

I was revising some differentiation, when I observed something that I pretty much always took for granted, so I decided to write it in mathematical notation. Would you please tell me if I'm correct to say the following:

If $f(X) = cg(X)$, $c$ is a constant then $f'(X) = cg'(x)$.

Where $f'(x)$ and $g'(x)$ means 'gradient of each function'.

• Yes, that's correct. – Cookie Mar 27 '15 at 22:25
• This follows directly from the definition: $f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\frac{cg(x+h)-cg(x)}{h} = c\lim_{h\to 0}\frac{g(x+h)-g(x)}{h} = cg'(x)$ using the fact that a constant factor commutes with the limit. – Winther Mar 27 '15 at 22:37
• @Winther: Post your comment as an answer...? :) – Andrew D. Hwang Mar 27 '15 at 23:00

This follows directly from the definition:

$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\to 0}\frac{cg(x+h)-cg(x)}{h} = c\lim_{h\to 0}\frac{g(x+h)-g(x)}{h} = cg'(x)$

using the fact that a constant factor commutes with the limit.

That usually one of the first theorems you learn about the derivative (right after 'the derivative of a constant function is 0').

You are correct.

let us have $h(x)=c$

$f(x)=c\cdot g(x)=h(x)\cdot g(x)$

Now using the product rule.

$f'(x)=h'(x)\cdot g(x)+g'(x)\cdot h(x)=0\cdot g(x) + g'(x)\cdot c=c\cdot g(x)$

Note that the derivative of a constant is zero.

• I think for this proof you shouldn't use the product rule, meaning that this result is known before the product rule is introduced. – Hirshy Aug 24 '15 at 6:21