1. We are given the function derivative expression: $$f'(x) = \sqrt{\arccos(3x+15) - \arccos(x+25)}$$. 2. To understand or work with this expression, we note the domain restrictions from arccos: the arguments must lie in $$[-1,1]$$. 3. Check domain for $$\arccos(3x+15)$$: $$-1 \leq 3x+15 \leq 1 \Rightarrow -16 \leq 3x \leq -14 \Rightarrow -\frac{16}{3} \leq x \leq -\frac{14}{3}$$. 4. Check domain for $$\arccos(x+25)$$: $$-1 \leq x+25 \leq 1 \Rightarrow -26 \leq x \leq -24$$. 5. The domain of the expression is intersection of these intervals: $$[-\frac{16}{3}, -\frac{14}{3}] \cap [-26, -24]$$ which is empty. 6. So there is no real $$x$$ satisfying both arguments in $$[-1,1]$$ simultaneously, making the expression undefined for real $$x$$. 7. If you want to differentiate an original function $$f(x)$$ which has derivative $$f'(x)$$ as above, you might need to reconsider the original function or verify the problem. Final answer: The derivative expression $$f'(x) = \sqrt{\arccos(3x+15) - \arccos(x+25)}$$ is not defined for any real $$x$$ due to domain restrictions of arccos.