1. **State the problem:** Find the derivative with respect to $x$ of the function $$f(x) = \ln \sqrt{x^2 + 19}.$$\n\n2. **Rewrite the function:** Recall that $$\sqrt{x^2 + 19} = (x^2 + 19)^{\frac{1}{2}}.$$ So, $$f(x) = \ln \left( (x^2 + 19)^{\frac{1}{2}} \right).$$\n\n3. **Use logarithm properties:** Using the property $$\ln(a^b) = b \ln a,$$ we get $$f(x) = \frac{1}{2} \ln (x^2 + 19).$$\n\n4. **Differentiate using the chain rule:** The derivative of $$\ln (x^2 + 19)$$ is $$\frac{1}{x^2 + 19} \cdot \frac{d}{dx}(x^2 + 19) = \frac{1}{x^2 + 19} \cdot 2x = \frac{2x}{x^2 + 19}.$$\n\n5. **Apply the constant multiple:** Since $$f(x) = \frac{1}{2} \ln (x^2 + 19),$$ its derivative is $$f'(x) = \frac{1}{2} \cdot \frac{2x}{x^2 + 19}.$$\n\n6. **Simplify the expression:** $$f'(x) = \frac{\cancel{1} \cdot \cancel{2} x}{\cancel{2} (x^2 + 19)} = \frac{x}{x^2 + 19}.$$\n\n**Final answer:** $$\frac{d}{dx} \left( \ln \sqrt{x^2 + 19} \right) = \frac{x}{x^2 + 19}.$$