Let $X$ and $Y$ be independent random variables with common distribution function $F$ and density function $f$. Show that $V = \max\{X,Y\}$ has distribution function $\textbf{P}(V \leq x) = F(x)^{2}$ and probability density function $f_{V}(x) = 2f(x)F(x), x\in\mathbb{R}$. Find the density function of $U = \min\{X,Y\}$.
MY ATTEMPT
Since $V = \max\{X,Y\}$, it results that \begin{align*} \textbf{P}(V\leq x) = \textbf{P}(\max\{X,Y\}\leq x) = \textbf{P}(X\leq x, Y\leq x) = \textbf{P}(X\leq x)\textbf{P}(Y\leq x) = F(x)^{2} \end{align*}
because $X$ and $Y$ are independent. Therefore \begin{align*} f_{V}(x) = \frac{\mathrm{d}}{\mathrm{d}x}F(x)^{2} = 2F(x)F^{\prime}(x) = 2f(x)F(x) \end{align*}
Finally, we have \begin{align*} \textbf{P}(U \leq x) & = \textbf{P}(\min\{X,Y\}\leq x) = 1 - \textbf{P}(\min\{X,Y\} > x)\\\\ & = 1 - \textbf{P}(X > x,Y > x) = 1 - \textbf{P}(X>x)\textbf{P}(Y>x)\\\\ & = 1 - (1 - F(x))(1 - F(x)) = F(x) + F(x) - F(x)F(x)\\\\ & = F(x)(2 - F(x)) \end{align*}
Could someone double-check my calculations? Thanks in advance!