Let $\lambda > 0$ and let $f(x)$ be a periodic function that has period $a$. How to show that function $g(x)=f'(\lambda x)$ is periodic and determine its period. Just some hints, please. I have achieved this far( I have used following function $g(x)=f(\lambda x)$): $\lambda > 0$ $g(x+a)=f(\lambda(x+a))=f(\lambda x+ \lambda a) = ...$
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Assume $f(t)=f(t+a)$. Then $f'(t)=f'(t+a)$.
Now $f'(\lambda x)=f'(\lambda x + a)$, taking $t=\lambda x$ from above.
I leave the last step for you. Since $g(x)=f'(\lambda x)$, what value can we add to $x$ in both the left and right expressions and still yield a true statement?
If $a$ is a period of $f$ (and $f$ is differentiable), then $a$ is clearly also a period of $f'$. Finally $\frac a\lambda$ is a period of $g$ as seen by plugging it in: $g(x+\frac a\lambda)=f'(\lambda x+a)=f'(\lambda x)=g(x)$.
Do you also need to show that $g$ has no period smaller than $\frac a\lambda$ if $a$ is the smallest period of $f$?