Showing $\frac{d\theta }{ d \tan \theta}=\frac{ 1}{ 1+ \tan^2 \theta}$ I suppose that
$$
\frac{d\theta }{ d \tan \theta}=\frac{d \arctan  x }{ d x}=  \frac{1}{1+x^2}=\frac{ 1}{ 1+ \tan^2 \theta} 
$$
So is
$$
\frac{d\theta }{ d \tan \theta}=\frac{ 1}{ 1+ \tan^2 \theta} 
$$
correct? And
$$
\frac{d (\theta)  }{ d \tan \frac{\theta}{2}}=\frac{ 2}{ 1+ \tan^2 \frac{\theta}{2}} \; ?
$$

Follows is the way I look at the derivation of
$\dfrac{d\theta}{d\tan \theta} = \dfrac{1}{1+ \tan^2 \theta} \tag 0$
and related identities; starting with
$\dfrac{d\theta}{d\tan \theta} = \dfrac{1}{\dfrac{d\tan \theta}{d\theta}} \tag 1$
we use the definition $\tan \theta = \sin \theta / \cos \theta$ and the quotient rule for derivatives to obtain:
$\dfrac{d\tan \theta}{d\theta} = \dfrac{d}{d\theta}\dfrac{\sin \theta}{\cos \theta} = \dfrac{(\cos \theta)(\cos \theta) - (-\sin \theta)(\sin \theta)}{\cos^2 \theta}$
$= \dfrac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} = \dfrac{\cos^2 \theta}{\cos^2 \theta} + \dfrac{\sin^2 \theta}{\cos^2 \theta} = 1+ \tan^2 \theta; \tag 2$
now by (1),
$\dfrac{d\theta}{d\tan \theta} = \dfrac{1}{1+ \tan^2 \theta}; \tag 3$
having obtained this result, we may calculate $\dfrac{d\theta}{d\tan (\theta/2)}$, additionally invoking the chain rule; we set
$u(\theta) = \dfrac{\theta}{2}, \tag4$
whence
$\dfrac{du(\theta)}{d\theta} = \dfrac{1}{2}, \tag 5$
and
$\dfrac{d\tan (\theta/2)}{d\theta} = \dfrac{d\tan u(\theta)}{d\theta} = \dfrac{d\tan u(\theta)}{du} \dfrac{du(\theta)}{d\theta}$
$=\dfrac{1}{2}(1 + \tan^2(u(\theta)) = \dfrac{1 + \tan^2 (\theta/2)}{2}, \tag 6$
and thus
$\dfrac{d\theta}{d\tan (\theta/2)} = \dfrac{2}{1 + \tan^2 (\theta/2)}. \tag 7$
$OE\Delta$.
