1. Problem: Differentiate the function $k(x)=x[\sin(\ln x)-\cos(\ln x)]$. 2. Formula: Use the product rule and the chain rule. 3. Product rule: $$ (fg)' = f' g + f g' $$ 4. Identify: Let $f(x)=x$ and $g(x)=\sin(\ln x)-\cos(\ln x)$. Then $f'(x)=1$ and the derivatives inside $g$ use the chain rule. 5. Differentiate the inner terms: $\dfrac{d}{dx}\sin(\ln x)=\cos(\ln x)\cdot\dfrac{1}{x}$ and $\dfrac{d}{dx}[-\cos(\ln x)]=\sin(\ln x)\cdot\dfrac{1}{x}$. 6. Apply the product rule: $$k'(x)=f'(x)g(x)+f(x)g'(x)=1\cdot[\sin(\ln x)-\cos(\ln x)]+x\cdot\frac{1}{x}[\cos(\ln x)+\sin(\ln x)].$$ 7. Simplify the second term by canceling $x$ with $\frac{1}{x}$: $$x\cdot\frac{1}{x}=\frac{\cancel{x}}{\cancel{x}}=1.$$ 8. Combine like terms: $$k'(x)=\sin(\ln x)-\cos(\ln x)+\cos(\ln x)+\sin(\ln x)=2\sin(\ln x).$$ 9. Final answer: $k'(x)=2\sin(\ln x)$.