1. **State the problem:** Find the derivative of the function $$y = \log_{10} \left( \frac{1+x}{1-x} \right)$$. 2. **Recall the formula:** The derivative of $$\log_a u$$ with respect to $$x$$ is $$\frac{1}{u \ln a} \cdot \frac{du}{dx}$$, where $$u$$ is a function of $$x$$ and $$a$$ is the base of the logarithm. 3. **Identify $$u$$:** Here, $$u = \frac{1+x}{1-x}$$. 4. **Differentiate $$u$$:** Using the quotient rule, $$$\frac{du}{dx} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^2} = \frac{(1-x) + (1+x)}{(1-x)^2} = \frac{2}{(1-x)^2}$$$ 5. **Apply the derivative formula:** $$$\frac{dy}{dx} = \frac{1}{u \ln 10} \cdot \frac{du}{dx} = \frac{1}{\frac{1+x}{1-x} \ln 10} \cdot \frac{2}{(1-x)^2} = \frac{2}{\ln 10} \cdot \frac{1-x}{1+x} \cdot \frac{1}{(1-x)^2} = \frac{2}{\ln 10} \cdot \frac{1}{(1+x)(1-x)} = \frac{2}{\ln 10 (1 - x^2)}$$$ 6. **Final answer:** $$$\boxed{\frac{dy}{dx} = \frac{2}{\ln 10 (1 - x^2)}}$$ This derivative tells us how the function changes with respect to $$x$$, using the chain rule and properties of logarithms.