Derivative Arctan Sqrt D315C8

1. **State the problem:** We need to find the derivative of the function $$y = \arctan\left(\sqrt{\frac{1 + x}{1 - x}}\right)$$. 2. **Recall the derivative formula:** The derivative of $$\arctan(u)$$ with respect to $$x$$ is $$\frac{u'}{1 + u^2}$$, where $$u$$ is a function of $$x$$. 3. **Identify the inner function:** Let $$u = \sqrt{\frac{1 + x}{1 - x}} = \left(\frac{1 + x}{1 - x}\right)^{\frac{1}{2}}$$. 4. **Find $$u'$$:** $$u' = \frac{1}{2} \left(\frac{1 + x}{1 - x}\right)^{-\frac{1}{2}} \cdot \frac{d}{dx} \left(\frac{1 + x}{1 - x}\right)$$ 5. **Differentiate the quotient inside:** $$\frac{d}{dx} \left(\frac{1 + x}{1 - x}\right) = \frac{(1 - x) \cdot 1 - (1 + x) \cdot (-1)}{(1 - x)^2} = \frac{1 - x + 1 + x}{(1 - x)^2} = \frac{2}{(1 - x)^2}$$ 6. **Substitute back into $$u'$$:** $$u' = \frac{1}{2} \left(\frac{1 + x}{1 - x}\right)^{-\frac{1}{2}} \cdot \frac{2}{(1 - x)^2} = \left(\frac{1 + x}{1 - x}\right)^{-\frac{1}{2}} \cdot \frac{1}{(1 - x)^2}$$ 7. **Simplify $$\left(\frac{1 + x}{1 - x}\right)^{-\frac{1}{2}}$$:** $$\left(\frac{1 + x}{1 - x}\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{\frac{1 + x}{1 - x}}} = \sqrt{\frac{1 - x}{1 + x}}$$ 8. **So, $$u'$$ becomes:** $$u' = \sqrt{\frac{1 - x}{1 + x}} \cdot \frac{1}{(1 - x)^2} = \frac{\sqrt{1 - x}}{\sqrt{1 + x} (1 - x)^2}$$ 9. **Now, compute $$y'$$:** $$y' = \frac{u'}{1 + u^2}$$ 10. **Calculate $$1 + u^2$$:** $$1 + u^2 = 1 + \left(\sqrt{\frac{1 + x}{1 - x}}\right)^2 = 1 + \frac{1 + x}{1 - x} = \frac{(1 - x) + (1 + x)}{1 - x} = \frac{2}{1 - x}$$ 11. **Substitute into $$y'$$:** $$y' = \frac{\frac{\sqrt{1 - x}}{\sqrt{1 + x} (1 - x)^2}}{\frac{2}{1 - x}} = \frac{\sqrt{1 - x}}{\sqrt{1 + x} (1 - x)^2} \cdot \frac{1 - x}{2}$$ 12. **Simplify:** $$y' = \frac{\sqrt{1 - x}}{\sqrt{1 + x} (1 - x)^2} \cdot \frac{1 - x}{2} = \frac{\sqrt{1 - x}}{\sqrt{1 + x} (1 - x)} \cdot \frac{1}{2} = \frac{1}{2} \cdot \frac{\sqrt{1 - x}}{\sqrt{1 + x} (1 - x)}$$ 13. **Simplify the fraction inside:** $$\frac{\sqrt{1 - x}}{(1 - x)} = \frac{1}{\sqrt{1 - x}}$$ 14. **Final derivative:** $$y' = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - x} \sqrt{1 + x}} = \frac{1}{2 \sqrt{1 - x^2}}$$ **Answer:** $$\boxed{y' = \frac{1}{2 \sqrt{1 - x^2}}}$$