Let be $f:(0, \frac{\pi}{6})\rightarrow R$, $f(x)=\tan(x)\tan(2x)\tan(3x)$. I need to prove that $f^{(n)}(x)>0$ for any $0 < x < \frac{\pi}{6}$, where $f^{(n)}(x)$ is the nth derivative of $f(x)$.
It seems rather cumbersome to derive it.
Let be $f:(0, \frac{\pi}{6})\rightarrow R$, $f(x)=\tan(x)\tan(2x)\tan(3x)$. I need to prove that $f^{(n)}(x)>0$ for any $0 < x < \frac{\pi}{6}$, where $f^{(n)}(x)$ is the nth derivative of $f(x)$.
It seems rather cumbersome to derive it.
Hint: The derivative of $\tan x$ is $1 + \tan^2 x$. This implies that your $n$th derivative will be a big polynomial in $\tan x$, $\tan 2x$ and $\tan 3x$ with only positive coefficients.