Question:

Step 1: Define the functions and apply the substitution. Let $u = \sin^{-1}\!\dfrac{2x}{1+x^2}$ and $v = \cos^{-1}\!\dfrac{1-x^2}{1+x^2}$. To find $\dfrac{du}{dv}$, we use the substitution $x = \tan\theta$, where $\theta \in (-\pi/2, \pi/2)$. Substituting $x=\tan\theta$ into the expressions for $u$ and $v$: $$ u = \sin^{-1}\!\left(\dfrac{2\tan\theta}{1+\tan^2\theta}\right) = \sin^{-1}(\sin 2\theta) $$ $$ v = \cos^{-1}\!\left(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \cos^{-1}(\cos 2\theta) $$ Step 2: Determine the expressions for $u$ and $v$ and calculate $\dfrac{du}{dv}$ based on the range of $x$. **Case 1: $x \in [0, 1)$** If $x \in [0, 1)$, then $\theta \in [0, \pi/4)$. This implies $2\theta \in [0, \pi/2)$. In this range, $u = \sin^{-1}(\sin 2\theta) = 2\theta$ and $v = \cos^{-1}(\cos 2\theta) = 2\theta$. Substituting back $\theta = \tan^{-1}x$: $u = 2\tan^{-1}x$ $v = 2\tan^{-1}x$ Differentiating with respect to $x$: $\dfrac{du}{dx} = \dfrac{2}{1+x^2}$ $\dfrac{dv}{dx} = \dfrac{2}{1+x^2}$ Therefore, $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} = \dfrac{2/(1+x^2)}{2/(1+x^2)} = 1$. **Case 2: $x \in (-1, 0)$** If $x \in (-1, 0)$, then $\theta \in (-\pi/4, 0)$. This implies $2\theta \in (-\pi/2, 0)$. In this range, $u = \sin^{-1}(\sin 2\theta) = 2\theta$. For $v$, since $2\theta \in (-\pi/2, 0)$, $\cos 2\theta = \cos(-2\theta)$, and $-2\theta \in (0, \pi/2)$. So, $v = \cos^{-1}(\cos(-2\theta)) = -2\theta$. Substituting back $\theta = \tan^{-1}x$: $u = 2\tan^{-1}x$ $v = -2\tan^{-1}x$ Differentiating with respect to $x$: $\dfrac{du}{dx} = \dfrac{2}{1+x^2}$ $\dfrac{dv}{dx} = -\dfrac{2}{1+x^2}$ Therefore, $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} = \dfrac{2/(1+x^2)}{-2/(1+x^2)} = -1$. **Case 3: $x > 1$** If $x > 1$, then $\theta \in (\pi/4, \pi/2)$. This implies $2\theta \in (\pi/2, \pi)$. For $u$, since $2\theta \in (\pi/2, \pi)$, $\sin 2\theta = \sin(\pi - 2\theta)$, and $\pi - 2\theta \in (0, \pi/2)$. So, $u = \sin^{-1}(\sin(\pi - 2\theta)) = \pi - 2\theta$. For $v$, since $2\theta \in (\pi/2, \pi)$, this is within the principal range of $\cos^{-1}$. So, $v = \cos^{-1}(\cos 2\theta) = 2\theta$. Substituting back $\theta = \tan^{-1}x$: $u = \pi - 2\tan^{-1}x$ $v = 2\tan^{-1}x$ Differentiating with respect to $x$: $\dfrac{du}{dx} = -\dfrac{2}{1+x^2}$ $\dfrac{dv}{dx} = \dfrac{2}{1+x^2}$ Therefore, $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} = \dfrac{-2/(1+x^2)}{2/(1+x^2)} = -1$. **Case 4: $x < -1$** If $x < -1$, then $\theta \in (-\pi/2, -\pi/4)$. This implies $2\theta \in (-\pi, -\pi/2)$. For $u$, since $2\theta \in (-\pi, -\pi/2)$, $\sin 2\theta = \sin(2\theta + \pi)$, and $2\theta + \pi \in (0, \pi/2)$. So, $u = \sin^{-1}(\sin(2\theta + \pi)) = 2\theta + \pi$. For $v$, since $2\theta \in (-\pi, -\pi/2)$, $\cos 2\theta = \cos(-2\theta)$, and $-2\theta \in (\pi/2, \pi)$. So, $v = \cos^{-1}(\cos(-2\theta)) = -2\theta$. Substituting back $\theta = \tan^{-1}x$: $u = \pi + 2\tan^{-1}x$ $v = -2\tan^{-1}x$ Differentiating with respect to $x$: $\dfrac{du}{dx} = \dfrac{2}{1+x^2}$ $\dfrac{dv}{dx} = -\dfrac{2}{1+x^2}$ Therefore, $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} = \dfrac{2/(1+x^2)}{-2/(1+x^2)} = -1$. Step 3: Summarize the results. The differential coefficient of $\sin^{-1}\!\dfrac{2x}{1+x^2}$ with respect to $\cos^{-1}\!\dfrac{1-x^2}{1+x^2}$ is: $$ \dfrac{du}{dv} = \begin{cases} 1 & \text{if } x \in [0, 1) \\ -1 & \text{if } x \in (-\infty, 0) \setminus \{-1\} \cup (1, \infty) \end{cases} $$ The derivative does not exist at $x = -1, 0, 1$ due to the piecewise definitions of the functions.