<s>[INST] Least Common Denominator: $ \frac{\sqrt{x}}{x}+\frac{\ln\ x}{2\sqrt{x}} $ $$ \frac{\sqrt{x}}{x}+\frac{\ln \ x}{2\sqrt{x}} $$
I have tried combining these two fractions; however, I keep getting stuck. 
$$\frac{2\sqrt{x}}{2\sqrt{x}}\cdot\frac{\sqrt{x}}{x}+\frac{\ln\ x}{2\sqrt{x}}\cdot \frac{x}{x}$$
So that we get
$$\frac {2x}{2\sqrt{x}\cdot(x)}+\frac {x\ln\ x}{2\sqrt{x}\cdot(x)}$$
What am I doing wrong? The answer is $$ \frac{2+\ln\ x}{2\sqrt{x}}$$
 [/INST] The square root $\sqrt{x}$ seems to be the thing that bothers you: pretend it's not there! That is, write temporarily $\sqrt{x}=z$ so that $x=z^2$; then
$$
\frac{\sqrt{x}}{x}+\frac{\ln x}{2\sqrt{x}}
=
\frac{z}{z^2}+\frac{\ln(z^2)}{2z}
$$
Now, for avoiding misunderstandings, write temporarily $\ln(z^2)=\ln x=L$ so you have
$$
\frac{z}{z^2}+\frac{L}{2z}=\frac{1}{z}+\frac{L}{2z}=
\frac{2+L}{2z}
$$
Substitute back to reinstate $x$:
$$
\frac{2+L}{2z}=\frac{2+\ln x}{2\sqrt{x}}
$$
This is longer, but safer. With some experience, you'll recognize more easily that
$$
\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}
$$
but the method of temporarily renaming objects to make them simpler works in several other cases.
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