Differentiate Ln Sqrt 228499

1. **State the problem:** Differentiate the function $$y=\ln(2+\sqrt{x})^2$$ with respect to $x$. 2. **Rewrite the function:** Using the property of logarithms, $$\ln(a)^2 = (\ln a)^2$$, so we have $$y = (\ln(2+\sqrt{x}))^2$$. 3. **Use the chain rule:** Let $$u = \ln(2+\sqrt{x})$$, then $$y = u^2$$. 4. **Differentiate $y$ with respect to $u$:** $$\frac{dy}{du} = 2u$$. 5. **Differentiate $u$ with respect to $x$:** $$u = \ln(2+\sqrt{x})$$ $$\frac{du}{dx} = \frac{1}{2+\sqrt{x}} \cdot \frac{d}{dx}(2+\sqrt{x})$$ Since $$\frac{d}{dx}(2) = 0$$ and $$\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}$$, $$\frac{du}{dx} = \frac{1}{2+\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}(2+\sqrt{x})}$$. 6. **Apply the chain rule to find $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u \cdot \frac{1}{2\sqrt{x}(2+\sqrt{x})} = \frac{2 \ln(2+\sqrt{x})}{2\sqrt{x}(2+\sqrt{x})}$$ 7. **Simplify the expression:** $$\frac{dy}{dx} = \frac{\cancel{2} \ln(2+\sqrt{x})}{\cancel{2} \sqrt{x}(2+\sqrt{x})} = \frac{\ln(2+\sqrt{x})}{\sqrt{x}(2+\sqrt{x})}$$. **Final answer:** $$\boxed{\frac{dy}{dx} = \frac{\ln(2+\sqrt{x})}{\sqrt{x}(2+\sqrt{x})}}$$