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?
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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).
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
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?
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