Differentiation of $\arctan\, v$

Let^5.11 $y =\ \arctan v$; then $y =\ \tan y$.

Figure: The inverse tangent $\tan^{-1}\ x$ using SAGE.

Figure: The standard branch of $\arctan\ x$ using SAGE.

Differentiating with respect to $y$ by (XIV),

$$\frac{dv}{dy} =\ \sec^2 y;$$

therefore $\frac{dy}{dv} = \frac{1}{\sec^2 y}$, by (5.2). But since $v$ is a function of $x$, this may be substituted in the formula $\frac{dy}{dx} = \frac{dy}{dv} \cdot \frac{dv}{dx}$, by (5.1),giving

$$\frac{dy}{dx} = \frac{1}{\sec^2 y} \cdot \frac{dv}{dx} = \frac{1}{1 + v^2} \frac{dv}{dx},$$

(since $\sec^2y = 1 + \tan^2y = 1 + v^2$). Therefore

$$\frac{d}{dx} (\arctan\, v) = \frac{\frac{dv}{dx}}{1 + v^2}$$

(equation (XX) in §5.1 above).