Answer
Yes, see explanations.
Work Step by Step
Step 1. Test if the function is continuous at $x=0$: $\lim_{x\to0}\frac{1-cos(x)}{x}=\lim_{x\to0}\frac{2sin^2(x/2)}{x}=\lim_{x\to0}\frac{sin(x/2)}{x/2}\frac{sin(x/2)}{1}=1\times0=0$. Thus the function is continuous at $x=0$.
Step 2. Test if the function is differentiable at $x=0$: $\lim_{x\to0}f'(x)=\lim_{x\to0}\frac{x\ sin(x)-(1-cos(x))}{x^2}=\lim_{x\to0}\frac{x\ sin(x)-(2sin^2(x/2))}{x^2}=\lim_{x\to0}\frac{x\ sin(x)}{x^2}-\lim_{x\to0}\frac{2sin^2(x/2)}{x^2}=\lim_{x\to0}\frac{\ sin(x)}{x}-\frac{1}{2}\lim_{x\to0}(\frac{sin(x/2)}{x/2})^2=1-\frac{1}{2}=\frac{1}{2}$
Step 3. We conclude that the function has a derivative at $x=0$ with $f'(0)=\frac{1}{2}$
