$\pi (x) = -1 + \pi(\sqrt x) + \sum \mu (d)\lfloor \frac {x}{d}\rfloor$, 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.

This should be Legendre’s formula for computing $\pi(x)$:
$$
\pi(x)=-1+\pi(\sqrt{x})+\lfloor x \rfloor-\sum_{p_i\le a}\left\lfloor \dfrac{ x }{(p_i)}\right\rfloor+\sum_{p_i<p_j\le a}\left\lfloor\dfrac{ x}{(p_ip_j)}\right\rfloor-\sum_{p_i<p_j<p_k\le a}\left\lfloor \dfrac{x}{(p_ip_jp_k)}\right\rfloor+\dots
$$
This is proved there.
