# PDF given distribution function

I am given a distribution $$F(x)=\frac{1}{2}+\frac{x}{2(1+|x|)}$$ and I need to find the pdf. I did the following-

$$f(x)=F'(x)=\frac{2(1+|x|).1 -2.sgn(x).x}{2(1+|x|)^2}=\frac{2+2|x|-2|x|}{2(1+|x|)^2}=\frac{1}{(1+|x|)^2}$$

Is this correct?

As you apply the quotient rule on $$\frac{u}{v}$$, you have identify $$v$$ to be $$2(1+|x|)$$ but you did not square the $$2$$.
Notice that $$\int_{-\infty}^\infty \frac{1}{(1+|x|)^2}\, dx=2.$$
Upon squaring the $$2$$ as well in the denominator, you should obtain the pdf to be $$\frac1{2(1+|x|)^2}$$.