TITLE: Sketch the function $y = {1 \over {{x^2}}}\ln x$
QUESTION [5 upvotes]: I don't know where to begin with this, the ${1 \over {{x^2}}}$ part of the function throws me off, how to do I go about this? How does one generally approach a question like this?

REPLY [3 votes]: The function is only defined for positive $x$. For $0<x\lt 1$, $\ln x$ is negative, large negative near $x=0$. Dividing by $x^2$ makes things much larger negative.
For $x$ large positive, the $x^2$ at the bottom crushes puny $\ln x$. So for $x$ large, our function is positive but close to $0$.
We might note that $\frac{\ln x}{x^2}$ is $0$ at $x=1$, and positive for $x\gt 1$.
For more detail, we compute the derivative.  This is $\frac{1-2\ln x}{x^3}$. Thus the derivative is $0$ at $x=e^{1/2}$, positive for $x\lt e^{1/2}$, and negative for $x\gt e^{1/2}$. 
So our function reaches a maximum at $x=e^{1/2}$. The maximum value is $\frac{1}{2e}$.
Now we have enough information to draw a pretty good sketch. For fine detail about concavity, one can look at the second derivative. If you do that calculation, you will find that the curve is facing down up to $x=e^{5/6}$, and then facing up. This does not add much in the way of useful information in this case. However, an exam question might ask about concavity, since it is not too hard to deal with in this example.