Si $$t=\tan\left(\frac{x}{2}\right)$$ entonces por regla de la cadena:
$$\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{2}\tan'\left(\frac{x}{2}\right)$$
La derivada de la tangente la podés hallar escribiendo $$\tan(x)=\left(\frac{\sin(x)}{\cos(x)}\right)$$:
$$\tan'(x)=\left(\frac{\sin(x)}{\cos(x)}\right)'=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=1+\left(\frac{\sin(x)}{\cos(x)}\right)^2=1+\tan^2(x)$$
Entonces:
$$\frac{\mathrm{d}t}{\mathrm{d}x}=\frac{1}{2}\left[1+\tan^2\left(\frac{x}{2}\right)\right]=\frac{1}{2}\left(1+t^2\right)$$
Saludos