I'm trying to work with the following sum: $$f:=\sum_{d\leq x}\mu(d)\tau(d) \Big[ \frac{x}{d}\Big] $$ Where $\mu$ is the Mobius function, $\tau(n)$ is the number of positive divisors of $n$ and $h(x)=[x]$ is the floor function.

We know that $$\sum_{d\leq x} \mu(d)\Big[ \frac{x}{d}\Big]=1,$$ $$ \sum_{d\leq x} \Big[ \frac{x}{d} \Big] \sim x\log(x),$$ $$\sum_{x\leq d} \frac{x}{d} \sim x\ln(x), $$ but what happens if we throw $\tau$ into the mix? Is it similar to $$g:=\sum_{d\leq x}\mu(d)\tau(d)\frac{x}{d}?$$ Or is it similar to multiple of $g$? I've tried Mobius inversion with a few different declarations of $f,g$ but either I'm simply not seeing it, or it doesn't work. Any ideas?