Monotonicity of $f(x) =\sin(\ln(x))-\cos(\ln(x))$ Find the interval in which $f(x) =\sin(\ln(x))-\cos(\ln(x))$ is increasing. 
After differentiating we get $$f'(x) = \frac{\cos\left(\ln(x)\right)}{x} +\frac{\sin\left(\ln(x)\right)}{x}$$
Now how do we analyze this expression?

$$f'(x)=\frac{\sin \log x+\cos \log x}{x}$$
$\log x$ is an increasing function. Substitute $u=\log x$. Denominator is positive because $x$ is argument of logarithm.
$\sin u + \cos u > 0\to \sin u > - \cos u$
$\sin u = -\cos u$ when $\tan u=-1$ when $u=\dfrac{3\pi}{4}+k\pi,\forall k\in\mathbb{Z}$
therefore $\sin u > - \cos u$ for 
$2k\pi<u<\dfrac{3\pi}{4}+2k\pi\lor \dfrac{5\pi}{4}+2k\pi<u<2(k+1)\pi$
that is
$2k\pi<\log x<\dfrac{3\pi}{4}+2k\pi\lor \dfrac{5\pi}{4}+2k\pi<\log x<2(k+1)\pi$
and finally
$e^{2k\pi}< x<e^{\frac{3\pi}{4}+2k\pi}\lor e^{\frac{5\pi}{4}+2k\pi}<x<e^{2(k+1)\pi}$
these intervals are very huge, for instance for $k=3$ the derivative is positive in 
$(1.5\times 10^8,1.62009\times 10^9)\cup (7.79343\times 10^9,8.22263\times 10^{10})$
and graph is pretty weird, too
Hope this helps
edit
$f''(x)=-\dfrac{2 \sin (\log (x))}{x^2}$
Thus $x=e^{2 \pi  k}$ are inflexion points because $f''(x)=0$
$x=e^{2 \pi  k+\frac{3 \pi }{4}}$ are maxima because $f''(x)<0$
$x=e^{2 \pi  k+\frac{5 \pi }{4}}$ are minima
as $k\to -\infty$ maxima and minima oscillate in intervals exponentially smaller and smaller when $x\to 0^+$.

