Q.Differentiate $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$ w.r.t. $\cos^{-1}\dfrac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}$

$u$: as shown in II6A (same function), $u=\dfrac12\tan^{-1}x$, so $\dfrac{du}{dx}=\dfrac{1}{2(1+x^2)}$.

$v$: $v=\cos^{-1}\!\left(\dfrac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}\right)$. Let $s=\sqrt{1+x^2}$ and $z=\dfrac{1+s}{2s}=\dfrac12+\dfrac{1}{2s}$. Then $\dfrac{dz}{dx}=-\dfrac{1}{2s^2}\cdot\dfrac{ds}{dx}=-\dfrac{1}{2s^2}\cdot\dfrac{x}{s}=-\dfrac{x}{2s^3}$.

Compute $1-z^2=(1-z)(1+z)=\left(\dfrac12-\dfrac{1}{2s}\right)\left(\dfrac32+\dfrac{1}{2s}\right)=\dfrac{(s-1)(3s+1)}{4s^2}$. Expanding $(s-1)(3s+1)=3s^2-2s-1$, and using $s^2=1+x^2$: $3s^2-2s-1=3(1+x^2)-2s-1=3x^2+2-2s$. So

Therefore …

Reduce $u$ via the $x=\tan\theta$ substitution as in II6A; differentiate $v$ directly by the chain rule for $\cos^{-1}$ …

Assuming $v$ must also reduce to a clean multiple of $\sin^{-1}x$ or $\tan^{-1}x$ by analogy with $u$ (it does not — $\dfrac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}$ simplifies only to $\cos^2(\theta/2)$ under $x=\ …