Question:

Step 1: Simplify $g(x)$. Let $x = \sin\theta$ for $\theta \in (-\pi/2, \pi/2)$. Then $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = |\cos\theta|$. Since $\theta \in (-\pi/2, \pi/2)$, $\cos\theta > 0$, so $\sqrt{1-x^2} = \cos\theta$. Substitute $x=\sin\theta$ into the argument of $2\tan^{-1}$ in $g(x)$: $$ \dfrac{x}{1+\sqrt{1-x^2}} = \dfrac{\sin\theta}{1+\cos\theta} $$ Using the half-angle identities $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ and $1+\cos\theta = 2\cos^2(\theta/2)$: $$ \dfrac{\sin\theta}{1+\cos\theta} = \dfrac{2\sin(\theta/2)\cos(\theta/2)}{2\cos^2(\theta/2)} = \tan(\theta/2) $$ Since $\theta \in (-\pi/2, \pi/2)$, we have $\theta/2 \in (-\pi/4, \pi/4)$. In this interval, $\tan^{-1}(\tan y) = y$. Therefore, $$ 2\tan^{-1}\left(\dfrac{x}{1+\sqrt{1-x^2}}\right) = 2\tan^{-1}\left(\tan(\theta/2)\right) = 2(\theta/2) = \theta $$ So, $g(x)$ simplifies to: $$ g(x) = \cos(\theta) $$ Since $x=\sin\theta$, we have $\theta = \sin^{-1}x$. Thus, $$ g(x) = \cos(\sin^{-1}x) = \sqrt{1-x^2} $$ Step 2: Simplify $f(x)$. Let $x = \sin\theta$ for $\theta \in (-\pi/2, \pi/2)$. Substitute $x=\sin\theta$ into the argument of $\tan^{-1}$ in $f(x)$: $$ \dfrac{1-x^2/2}{x} = \dfrac{1-\sin^2\theta/2}{\sin\theta} $$ Using the identity $\sin^2\theta = \dfrac{1-\cos(2\theta)}{2}$: $$ \dfrac{1-\frac{1-\cos(2\theta)}{2}}{\sin\theta} = \dfrac{\frac{2-(1-\cos(2\theta))}{2}}{\sin\theta} = \dfrac{1+\cos(2\theta)}{2\sin\theta} $$ Using the identity $1+\cos(2\theta) = 2\cos^2\theta$: $$ \dfrac{2\cos^2\theta}{2\sin\theta} = \dfrac{\cos^2\theta}{\sin\theta} = \cot\theta $$ Since $\theta \in (-\pi/2, \pi/2)$, we have $\cot\theta = \tan(\pi/2-\theta)$. Therefore, $$ \tan^{-1}\left(\dfrac{1-x^2/2}{x}\right) = \tan^{-1}(\cot\theta) = \tan^{-1}(\tan(\pi/2-\theta)) $$ Since $\theta \in (-\pi/2, \pi/2)$, $\pi/2-\theta \in (0, \pi)$. For $\tan^{-1}(\tan y)$, if $y \in (0, \pi)$, then $\tan^{-1}(\tan y)$ is $y$ if $y \in (0, \pi/2)$ and $y-\pi$ if $y \in (\pi/2, \pi)$. For $x \in (0,1)$, $\theta \in (0, \pi/2)$, so $\pi/2-\theta \in (0, \pi/2)$. Thus, $\tan^{-1}(\tan(\pi/2-\theta)) = \pi/2-\theta$. So, $f(x)$ simplifies to: $$ f(x) = \sin(\pi/2-\theta) = \cos\theta $$ Since $x=\sin\theta$, we have $\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-x^2}$. $$ f(x) = \sqrt{1-x^2} $$ Step 3: Calculate $h(x)$. We have $f(x) = \sqrt{1-x^2}$ and $g(x) = \sqrt{1-x^2}$ for $x \in (0,1)$. $$ h(x) = f(x) - g(x) = \sqrt{1-x^2} - \sqrt{1-x^2} = 0 $$ Thus, $h(x)$ is a constant function for $x \in (0,1)$. Step 4: Calculate $h'(1/2)$. Since $h(x) = 0$ for $x \in (0,1)$, its derivative $h'(x)$ is also $0$ for $x \in (0,1)$. Therefore, $$ h'(1/2) = 0 $$