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?
 A: Just to rewrite your derivative in better latex
\begin{equation}
\frac{d}{dx}\Big(\sin(\ln(x))-\cos(\ln(x))\Big)=\frac{\cos(\ln(x))}{x}+\frac{\sin(\ln(x))}{x}.
\end{equation}
We want the interval in which the function is increasing, so we want the derivative to be positive
\begin{equation}
\frac{\cos(\ln(x))}{x}+\frac{\sin(\ln(x))}{x}>0.
\end{equation}
The variable x must be greater than zero for the log to be defined, so we get
\begin{equation}
\cos(\ln(x))+\sin(\ln(x))>0.
\end{equation}
To find the critical points, set the two terms equal
\begin{equation}
\cos(\ln(x))=-\sin(\ln(x)) \rightarrow \tan(\ln(x))=-1.
\end{equation}
This equality is satisfied for 
\begin{equation}
\ln(x)=-\frac{\pi}{4}+n\pi \rightarrow x=e^{-\pi/4+n\pi}.
\end{equation}
So the infinite number of intervals for which your function is increasing are seperated by the points $e^{-\pi/4+n\pi}$ for any integer $n$.  To see which one is increasing consider $x=1$ which is between the value $n=-1$ and $n=0$.  For this value the derivative is $1$ and so it is increasing.  The other intervals alternate between decreasing and increasing.
A: HINT:  Try applying the identity $\cos \theta + \sin \theta = \sqrt{2} \sin(\theta + \frac{\pi}{4} )$ to the derivative, or $\sin \theta - \cos \theta = \sqrt{2} \sin(\theta - \frac{\pi}{4})$ to the original function.
A: I like Michael Seifer's hint above.  It is a useful trick to have in your kit.
However, picking up where you left off.
$\frac{(\cos(\ln(x)))}{x} +\frac{(\sin(\ln(x)))}{x}>0$
$\ln x$ is not defined when $x \le 0$
$\cos(\ln(x)) +\sin(\ln(x))>0\\
\cos(\ln(x)) > - \sin(\ln(x))\\
\ln x \in (-\frac {\pi}{4} + 2n\pi, \frac {3\pi}{4} + 2n\pi)\\
x \in (e^{-\frac {\pi}{4} + 2n\pi}, e^{\frac {3\pi}{4} + 2n\pi})$
A: $$f'(x)=\frac{\sin \log x+\cos \log x}{x}$$
$\log x$ 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 $x\to 0^+$.

A: You can analyze the function $g(x)=\sin x-\cos x$ instead (see Seifert's answer). After that, assume that
$$e^a\le t<t'\le e^b$$
Then $ a\le\ln t<\ln t'\le b$. Therefore, if $g$ is increasing in $[a,b]$, we have $f(t)<f(t')$. Same reasoning if $g$ is decreasing in the interval.
Thus, $f$ is (in)(de)creasing in $[e^a,e^b]$ if and only if $g$ is (in)(de)creasing in $[a,b]$.
