1. **State the problem:** Simplify the derivative expression $$f'(x) = - \frac{\sqrt{1+x}}{(1+x)^2 \sqrt{1-x}}$$ 2. **Rewrite the denominator:** Notice that $(1+x)^2 = (1+x)(1+x)$, so we can write $$f'(x) = - \frac{\sqrt{1+x}}{(1+x)(1+x) \sqrt{1-x}}$$ 3. **Simplify the numerator and one factor in the denominator:** Since the numerator is $\sqrt{1+x}$ and one factor in the denominator is $(1+x)$, rewrite $(1+x)$ as $\sqrt{1+x} \cdot \sqrt{1+x}$: $$f'(x) = - \frac{\sqrt{1+x}}{\sqrt{1+x} \cdot \sqrt{1+x} \cdot (1+x) \sqrt{1-x}}$$ But this is redundant; instead, factor $\sqrt{1+x}$ in numerator and denominator: $$f'(x) = - \frac{\sqrt{1+x}}{(1+x)(1+x) \sqrt{1-x}} = - \frac{1}{(1+x) \sqrt{1-x} (1+x)/\sqrt{1+x}}$$ More straightforwardly, cancel one $\sqrt{1+x}$ from numerator and denominator: $$f'(x) = - \frac{1}{(1+x) \sqrt{1-x} (1+x)/\sqrt{1+x}}$$ 4. **Cancel $\sqrt{1+x}$:** Since $\sqrt{1+x} \cdot \sqrt{1+x} = 1+x$, we can cancel $\sqrt{1+x}$ in numerator and one $(1+x)$ in denominator: $$f'(x) = - \frac{1}{(1+x) \sqrt{1-x} \cdot (1+x)/\sqrt{1+x}} = - \frac{1}{(1+x) \sqrt{1-x} \cdot \sqrt{1+x}}$$ 5. **Combine the square roots in the denominator:** Multiply $\sqrt{1-x}$ and $\sqrt{1+x}$: $$\sqrt{1-x} \cdot \sqrt{1+x} = \sqrt{(1-x)(1+x)} = \sqrt{1 - x^2}$$ 6. **Final simplified form:** $$f'(x) = - \frac{1}{(1+x) \sqrt{1 - x^2}}$$ This matches the given simplified expression. **Summary:** We simplified by canceling $\sqrt{1+x}$ in numerator and denominator and combining the square roots in the denominator using the difference of squares identity.