$\pi (x) = -1 + \pi(\sqrt x) + \sum \mu (d)\lfloor \frac {x}{d}\rfloor$, Sorry about the title i didn't know what to call this other than "analytic number theory"
I am asked to show that $\pi (x) = -1 + \pi(\sqrt x) + \sum \mu (d)\lfloor \frac {x}{d}\rfloor$, where the sum is over all d for which ever prime factor is less or equal to $\sqrt x $
The furthest right hand part of the equation looks like a mobius inversion but im not sure how to do it over root x and the rest i am completely lost on.

I don't believe this is true. Take $x=25$. 
$\pi(x)=9$, $\pi(\sqrt{x})=3$.
The relevant $d$s are $2, 3, 5$ together with their square-free multiples, that is $6, 10, 15, 30$. The factor $\lfloor \frac {x}{d}\rfloor$ kills the participation of $15$ and $30$.
For the rest we get $$9=-1+3+(-12-8-5+4+2)$$
which is wrong, right?
