Remember that inverse "co-" functions often have the negative of their counterpart's derivative. For example, $\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2}$, which helps distinguish between options (A) and (D).
1. Setup: Let $y = \tan^{-1} x$. Then $\tan y = x$.
2. Differentiate Implicitly: Differentiate both sides with respect to $x$:
$$\sec^2 y \frac{dy}{dx} = 1$$
$$\frac{dy}{dx} = \frac{1}{\sec^2 y}$$
3. Convert back to $x$: Using the trigonometric identity $\sec^2 y = 1 + \tan^2 y$:
$$\frac{dy}{dx} = \frac{1}{1 + \tan^2 y}$$
Since we established that $\tan y = x$:
$$\frac{dy}{dx} = \frac{1}{1 + x^2}$$
This is a fundamental derivative used frequently in calculus and integration.