1. **Problem:** Find the derivative of $$y = x \sqrt{\frac{x^2 - 1}{x^2 + 2x + 1}}$$. 2. **Step 1: Simplify the expression inside the square root.** Note that $$x^2 + 2x + 1 = (x+1)^2$$. So, $$y = x \sqrt{\frac{x^2 - 1}{(x+1)^2}} = x \cdot \frac{\sqrt{x^2 - 1}}{|x+1|}$$. 3. **Step 2: Write the function as a product to apply the product rule.** Let $$u = x$$ $$v = \frac{\sqrt{x^2 - 1}}{|x+1|}$$ 4. **Step 3: Differentiate using the product rule:** $$y' = u'v + uv'$$ where $$u' = 1$$ 5. **Step 4: Differentiate $$v$$.** Rewrite $$v$$ as $$v = \frac{(x^2 - 1)^{1/2}}{|x+1|}$$. 6. **Step 5: Use the quotient rule for $$v$$:** $$v = \frac{f}{g}$$ with $$f = (x^2 - 1)^{1/2}, \quad g = |x+1|$$. 7. **Step 6: Compute $$f'$$ and $$g'$$:** $$f' = \frac{1}{2}(x^2 - 1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}}$$ For $$g = |x+1|$$, derivative is $$g' = \frac{d}{dx} |x+1| = \frac{x+1}{|x+1|}$$ (sign function). 8. **Step 7: Apply quotient rule:** $$v' = \frac{f'g - fg'}{g^2} = \frac{\frac{x}{\sqrt{x^2 - 1}} \cdot |x+1| - (x^2 - 1)^{1/2} \cdot \frac{x+1}{|x+1|}}{(x+1)^2}$$ 9. **Step 8: Simplify numerator:** $$= \frac{x|x+1|}{\sqrt{x^2 - 1}} - \frac{(x^2 - 1)^{1/2}(x+1)}{|x+1|}$$ Since $$|x+1|$$ cancels with denominator in second term, $$= \frac{x|x+1|}{\sqrt{x^2 - 1}} - \frac{(x^2 - 1)^{1/2}(x+1)}{|x+1|}$$ 10. **Step 9: Combine terms carefully and substitute back into $$y'$$:** $$y' = 1 \cdot v + x \cdot v' = v + x v'$$ 11. **Final answer:** $$y' = \frac{\sqrt{x^2 - 1}}{|x+1|} + x \cdot \frac{\frac{x}{\sqrt{x^2 - 1}} |x+1| - \frac{(x^2 - 1)^{1/2} (x+1)}{|x+1|}}{(x+1)^2}$$ This derivative expression can be further simplified depending on the domain of $$x$$ and sign of $$x+1$$. --- **Note:** The user asked multiple questions, but per instructions, only the first problem is solved here.