Given an arbitrary composite elementary function, does there exist some method to give us an intuition of its graph? Consider this example from the wikipedia page on elementary functions:
$$ \ \ \frac{e^{\tan x}}{1+x^2}\sin(\sqrt{1+(\ln x)^2})$$
Although one might know how any single simple elementary function that constitutes the composite function behaves and what its graph looks like, once you compose two or three elementary functions, it becomes hard to gain an intuition of how they behave, unless we use a computer.
Are there any standard methods one can use to, maybe, decompose them to make them easier to visualize? Or any method that could give us an idea of the graph, without actually computing it? Or is there absolutely no way to escape a case-by-case basis? Finding the extrema is the only method I know of that helps us in this direction.
 A: I decided to write up a more complete answer to this question. Clearly, this must be treated on a case-by-case basis, to an extent. However, often certain approximations can be made to simplify the problem. I will go thoroughly through the example given, and explain the thought process behind the approximations made. The function given as an example was
$$ \ \ \frac{e^{\tan x}}{1+x^2}\sin(\sqrt{1+(\ln x)^2}).$$
Let us take a look at the compositions, and try to make first-level approximations of them. We begin with the exponential. $e^{w}$ will be far larger that $1$ for $w>>0$, and is fairly close to $0$ when $w<<0$. $w=\tan x >0$ when $x\in (0,\pi/2)$, and $<0$ when $x\in (-\pi/2,0)$, and this is then appended. Thus, it is clear that when $x\in (0,\pi/2), e^{\tan x}$ will begin at $1$ and approach infinity in a fashion very similar to $\tan x$, perhaps a little faster. $\tan x$ gets very negative very fast as $x\rightarrow -\pi/2$ from the right. Thus, $e^{\tan x}\approx0 $ when $x\in(-\pi/2,0)$ and $\approx\tan x$ when $x\in(0,\pi/2)$; we call this "the approximation". We can then append this result. For the other composition, we note that $\sin y$ will lie in the interval $[-1,1]$. Multiplying our approximation of $e^{\tan x}$ by $\sin y$ will have almost no affect on the magnitude of $e^{\tan x}$, except possibly at $x=0$. However, we will need to study when $\sin y$ is greater/less than $0$ to determine if it flips the sign of the approximation.
The term $1+x^2$ is relatively innocent in the denominator; clearly $x^2$ wont compete with the asymptote at $x=\pi/2$, and there is no singularity at $0$. It will, however, make the approximation curve more sharp, especially for the appended intervals further from zero.
Lastly, we hone in on the sign of the approximation. The term $\sin(\sqrt{1+(\ln x)^2})$ is what causes it. We should now note that $\ln v$ is not defined when $v\leq0$, so the entire function isn't defined when $x\in(-\infty,0]$; of course, we still append the approximation, but don't plot anything to left of axis. We need to check how oscillatory our $\sin$ is; does it change sign several times in a single interval of the approximation? $\sin y$ is positive when $y\in(0,\pi)$, and negative when $y\in(\pi,2\pi)$, so a sign change occurs every $\pi$. Close to $0$, $\ln$ is hard to predict since it is very large, and thus will oscillate very quickly. However, for $x>1$, it is much easier, as $\ln$ slows down pretty quickly and continuously increases. At $x=1$, we have $\sin(1)$ which is a very small positive number. The $n$th sign change occurs when
$$\sqrt{1+(\ln x)^2}=n\pi$$
which is when $x=\exp[\sqrt{n^2\pi^2-1}]$. Then the first sign change is at around $20$ and the distance to the next sign change becomes exponentially further.


Ending remarks: The example provided could have been much harder; imagine oscillations within oscillations, and if the oscillation are too fast, there is almost nothing you can do unless they are very small relative to a term added e.g. $e^x+\sin x$. However, I hope this gives the impression that these are somewhat less scary than they seem, at least if they are elementary functions.
A: CURVE SKETCHING IN $4$ STEPS:

*

*STEP ($y$-info) The following are found if exist: * Domain * Range * y-intercept * x-intercepts * Asymptotes (Vertical, horizantal, oblique) * Limits at asymptotes and at bad points * Symmetries


*STEP ($y'$ -info) Sign table of $y'$ is made and the following are found if exist: * Increase/Decrease intervals * Local maximum and minum points


*STEP ($y''$-info) Sign table of $y''$ is made and the following are found if exist: * Concave Up/Down intervals * Inflection points


*STEP Sketching: Mark all points found on a Cartesian plane. Jagged-draw the aymptotes. Draw the curve obeying increase/decrease, concavity and the tangency at the asymptotes considering limits.
This is a version of a short reminder I wrote in my lecture notes many years ago for the curves whose first and second derivatives are easy to find and easy to analyze. Of course, each bullet item must be explained in detail which is done in calculus books.
Unfortunately, when the curve is too complicated, we should skip Steps 2 and 3 and get as much as we can from only Step 1. In fact, many curves can be sketched with Step 1 by finding as many test points as possible which is the most primitive way to draw a thing. I think O.P. is interested in Step 1 since the curve given is $y=f(x)=\frac{e^{\tan x}}{1+x^2}\sin\sqrt{1+\ln^2x}$.
Let me try to do Step 1, roughly in my style for the given function with help of WA. * Domain: $\ln x$ is defined for $x>0$, $\tan x$ is defined when $x\neq\frac{\pi}{2}+k\pi$, $k\in\Bbb{Z}$. So, $D=\{x|x>0,\; x\neq\frac{\pi}{2}+k\pi,\; k\in\Bbb{Z}\}$. * Range: All numbers (Why?) * $y$-intercept: None. * $x$-intercepts: $f(x)=0$ gives $\sin\sqrt{1+\ln^2x}=0$, hence $x=\exp(\pm\sqrt{n^2\pi^2-1})$ where $n\in\Bbb{Z}^+$. * Asymptotes: $\sin x$ is bounded, $x^2+1$ is never zero, so all vertical asymptotes are found from $\tan x=\pm\infty$ and they are $x=\frac{\pi}{2}+k\pi$, $k\in\Bbb{N}$, where the function is undefined. We notice that the limit from left to these asymptotes is $\pm\infty$ depending on the location of the zeros of the function: $+\infty$, up until $e^{\sqrt{\pi^2-1}}\approx 19.652$ with 5 asymptotes, then $-\infty$, down until  $e^{\sqrt{4\pi^2-1}}\approx 494.277$ with 151 asymptotes, etc. But limit from right to asymptotes is zero since $e^{-\infty}=0$. There is no horizontal asymptote in my opinion altough WA said limit at infnity is zero. * Limits at asymptotes and at bad points: We said that except zero. Vertical asymptotes are bad points and limit to them from right is zero. $x=0$ is a very bad point. The function oscillates from right towards to it taking values between numbers almost $1$ and almost $-1$. When the curve is oscillating, it cuts the $x$-axis at the small zeros which are $\exp(-\sqrt{n^2\pi^2-1})$ where $n\in\Bbb{Z}^+$. The first and the largest of these is $e^{-\sqrt{\pi-1}}\approx 0.0509$. * Symmetries: None. I hope I add something later here. I know there are lots of things to say for this curve. Each piece of the curve is interesting. And Step 2 may be tried...
