Is $\sum_{\rho}\frac{x^{\rho+1}}{\rho\left(\rho+1\right)}$ differentiable term by term?

Question

Let $\rho$ be the non trivial zeros of the Riemann zeta function and $x>1$. I would like to prove (or disprove) that $$\frac{d}{dx}\left(\sum_{\rho}\frac{x^{\rho+1}}{\rho\left(\rho+1\right)}\right)=\sum_{\rho}\frac{x^{\rho}}{\rho}.\tag{1}$$ I know that the series converges absolutely $$\sum_{\rho}\left|\frac{x^{\rho+1}}{\rho\left(\rho+1\right)}\right|\leq x^{2}\sum_{\rho}\left|\frac{1}{\rho\left(\rho+1\right)}\right|<+\infty$$ but I don't see how to prove that the series converges uniformly. I also know that $\sum_{\rho}\frac{x^{\rho+1}}{\rho\left(\rho+1\right)}$ is linked to the function $$\psi_{1}\left(x\right)=\int_{0}^{x}\psi\left(t\right)dt$$ where $\psi\left(t\right)$ is the Chebyshev psi function but I'm not sure if it can helps. Is there a way to prove (or disprove) $(1)$?