1. **State the problem:** Differentiate the function $$y = (x^2 + 1) \tan^{-1}(x) - x$$ and simplify the result. 2. **Recall the formula:** The derivative of $$\tan^{-1}(x)$$ is $$\frac{1}{1+x^2}$$. 3. **Apply the product rule:** For $$y = u v - x$$ where $$u = x^2 + 1$$ and $$v = \tan^{-1}(x)$$, the derivative is $$ \frac{dy}{dx} = u'v + uv' - 1 $$ where $$u' = 2x$$ and $$v' = \frac{1}{1+x^2}$$. 4. **Calculate each term:** $$ \frac{dy}{dx} = 2x \tan^{-1}(x) + (x^2 + 1) \frac{1}{1+x^2} - 1 $$ 5. **Simplify the fraction:** $$ (x^2 + 1) \frac{1}{1+x^2} = \frac{x^2 + 1}{1 + x^2} $$ Since numerator and denominator are the same, cancel common factors: $$ \frac{\cancel{x^2 + 1}}{\cancel{1 + x^2}} = 1 $$ 6. **Final simplified derivative:** $$ \frac{dy}{dx} = 2x \tan^{-1}(x) + 1 - 1 = 2x \tan^{-1}(x) $$ **Answer:** $$\boxed{\frac{dy}{dx} = 2x \tan^{-1}(x)}$$