How can I show a decreasing derivative? I have a derivative
$f'(x) = -1 + 8 \cos \frac{1}{x} + 4 \sin \frac{1}{x}$ 
And we have to show that it's decreasing in all intervals on the form $[\frac {6}{(12 n + 11) \pi } , \frac {6}{(12n + 7)\pi}$ for n>= 1
Our hint is to use the derivative such as $\frac {6}{(12n + 7)\pi}  <= \frac{6}{19 \pi}  < \frac{1}{8}$ 
I have no idea what to do other than 
$f'(x) = - 1 + 8 \cos \frac{1}{x} + 4 \sin \frac{1}{x} < \frac {6}{(12 n + 11) \pi } $ 
Please help. And thank you.

Hint: $$x \in \left[\frac {6}{(12 n + 11) \pi } , \frac {6}{(12n + 7)\pi}\right] \iff \frac1x \in \left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$$

 We have $x,y \in \left[\frac {6}{(12 n + 11) \pi },\frac {6}{(12n + 7)\pi}\right]$ and $x<y$ iff $\frac1x>\frac1y$ in $\left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$.  So this question can be rewritten as proving that $g(x) = 1+8\cos x+4\sin x$ is increasing in $\left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$.  Since $\sin$ and $\cos$ are periodic with period $2\pi$, the "$2n$"'s in the previous interval are redundant.  You may continue from here.  By either calculating $g'(x)$ OR rewriting the two trigonometric terms as one single term with a phase change.

