# Thread: k constant in first principle formula

1. ## k constant in first principle formula

Let f be a differentiable function. For any constant k > 0, put g(x) = f(kx). Prove from first principles that g'(a) = kf'(ka) for all a. Can you explain what this result means geometrically?

2. I presume that by "first principles" you mean using $\displaystyle g'(a)= \lim_{h\to 0}\frac{g(a+h)- g(a)}{h}$.

Just replace g(a) by kf(a).

3. I got:

lim
h -> 0 kf(a + h) - kf(a) / h

lim
h -> 0 k[f(a + h) - kf(a)] / h

I dont see a way to eliminate the h

4. Originally Posted by TsAmE
I got:

lim
h -> 0 kf(a + h) - kf(a) / h

lim
h -> 0 k[f(a + h) - kf(a)] / h

I dont see a way to eliminate the h
You don't! That is
$\displaystyle k\left(\lim_{h\to 0}\frac{f(a+h)- f(a)}{h}\right)$.

Now, what is that equal to?