Given the cumulative distribution $F(x)=\frac{x}{x+1}$ if $x\ge 0$, otherwise $F(x)=0$.
I calculated the probability density by doing the derivative of $\frac{x}{x+1}$ which is $\frac{1}{(x+1)^2}$ but I don't know if the probability density is $\frac{1}{(x+1)^2}$ when $x\ge 0$ and $0$ otherwise, or it's $\frac{1}{(x+1)^2}$ when $x>0$ and $0$ otherwise. My textbook says it's $\frac{1}{(x+1)^2}$ when $x\ge 0$ and $0$ otherwise but I don't understand why.