1. The problem is to find the derivative of the function given as $$f'(x) = 9\sqrt{x + 5} - 9\sqrt{x - 1}$$. 2. Recall the derivative rule for a square root function: $$\frac{d}{dx} \sqrt{u} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx}$$. 3. We will differentiate each term separately. 4. For the first term, let $$u = x + 5$$, so $$\frac{du}{dx} = 1$$. 5. The derivative of $$9\sqrt{x + 5}$$ is $$9 \cdot \frac{1}{2\sqrt{x + 5}} \cdot 1 = \frac{9}{2\sqrt{x + 5}}$$. 6. For the second term, let $$v = x - 1$$, so $$\frac{dv}{dx} = 1$$. 7. The derivative of $$9\sqrt{x - 1}$$ is $$9 \cdot \frac{1}{2\sqrt{x - 1}} \cdot 1 = \frac{9}{2\sqrt{x - 1}}$$. 8. Since the original function is $$f(x) = 9\sqrt{x + 5} - 9\sqrt{x - 1}$$, its derivative is: $$f'(x) = \frac{9}{2\sqrt{x + 5}} - \frac{9}{2\sqrt{x - 1}}$$. 9. This is the simplified form of the derivative. Final answer: $$f'(x) = \frac{9}{2\sqrt{x + 5}} - \frac{9}{2\sqrt{x - 1}}$$.