1. **Problem:** Find the derivative of $y = \tan^{-1}(1)$ and identify which expression matches the derivative. 2. **Recall the derivative formula:** The derivative of $y = \tan^{-1}(x)$ is given by $$y' = \frac{1}{1 + x^2}$$ 3. **Evaluate at $x=1$:** Since $y = \tan^{-1}(1)$ is a constant value (the arctangent of 1), its derivative with respect to $x$ is 0. 4. **Check the options:** None of the options directly show 0, but the derivative formula for $\tan^{-1}(x)$ is $\frac{1}{1+x^2}$, so the derivative at $x=1$ is $$y' = \frac{1}{1 + 1^2} = \frac{1}{2}$$ 5. **Conclusion:** The derivative of $y = \tan^{-1}(x)$ is $\frac{1}{1+x^2}$, so the correct general derivative expression is option 3: $$y' = - \frac{1}{1 + x^2}$$ However, the negative sign is incorrect for $\tan^{-1}(x)$ derivative, so the correct derivative is positive $\frac{1}{1+x^2}$. Since none of the options exactly match $\frac{1}{1+x^2}$, the closest correct formula is option 3 but without the negative sign. **Final answer:** The derivative of $y = \tan^{-1}(x)$ is $$y' = \frac{1}{1 + x^2}$$ --- **Slug:** arctan derivative **Subject:** calculus **Desmos:** {"latex":"y=\tan^{-1}(x)","features":{"intercepts":true,"extrema":true}} **q_count:** 6