$\sum_{n=1}^{+\infty}\frac{\Lambda\left(n\right)\varphi\left(n\right)}{n^{s}}$ and Riemann Zeta function

Question

Is it possible to write $\sum_{n=1}^{+\infty}\frac{\Lambda\left(n\right)\varphi\left(n\right)}{n^{s}}$, where $\Lambda(n)$ is the Von Mangoldt function and $\varphi(n)$ is the Euler totient function, in terms of the Riemann Zeta function? I know that $\sum_{n=1}^{+\infty}\frac{\varphi\left(n\right)}{n^{s}}=\frac{\zeta\left(s-1\right)}{\zeta\left(s\right)}=:F(s)$ but since $\varphi\left(n\right)$ is not completely multiplicative, we cannot conclude that the function is $F'(s)/F(s)$. I also found this result but I don't know if it is useful or not.

Answer 1

For $\Re(s)> 0$

$$\sum_p p^{-s} \log p = \sum_{d\ge 1} \mu(d) \frac{-\zeta'(sd)}{\zeta(sd)}$$ And you are looking at $$\sum_{n\ge 1} \Lambda(n) \varphi(n)n^{-s}= \sum_{k\ge 1} \sum_p (p^{k(1-s)}-p^{k(1-s)-1})\log p $$