Showing that $\frac{d}{dx}ln(u_{1}(x)\cdot ... \cdot u_{N}(x))=\sum_{j=1}^{N} \frac{u_{j}'(x)}{u_{j}(x)}$

Question

$\frac{d}{dx}ln(u_{1}(x)\cdot ... \cdot u_{N}(x))=\sum_{j=1}^{N} \frac{u_{j}'(x)}{u_{j}(x)}$ given that $u_{j}:\mathbb{R}\rightarrow(0,\infty)$ and $1\leq j \leq N$

Answer 1

HINT:

$$\ln(fg)=\ln(f)+\ln(g)$$

$$\frac d{dx}f+g=f'+g'$$

Can you try using these rules?