# Question #1679b

Jun 30, 2017

$\frac{d}{\mathrm{dx}} \tan \left(\frac{x}{2}\right) = \frac{1}{2} {\sec}^{2} \left(\frac{x}{2}\right)$

#### Explanation:

By definition:

$\frac{d}{\mathrm{dx}} \tan \left(\frac{x}{2}\right) = {\lim}_{h \to 0} \frac{\tan \left(\frac{x + h}{2}\right) - \tan \left(\frac{x}{2}\right)}{h}$

Using the trigonometric formula:

$\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$

we get:

$\frac{d}{\mathrm{dx}} \tan \left(\frac{x}{2}\right) = {\lim}_{h \to 0} \frac{\frac{\tan \left(\frac{x}{2}\right) + \tan \left(\frac{h}{2}\right)}{1 - \tan \left(\frac{x}{2}\right) \tan \left(\frac{h}{2}\right)} - \tan \left(\frac{x}{2}\right)}{h}$

$\frac{d}{\mathrm{dx}} \tan \left(\frac{x}{2}\right) = {\lim}_{h \to 0} \frac{1}{h} \left(\frac{\tan \left(\frac{x}{2}\right) + \tan \left(\frac{h}{2}\right) - \tan \left(\frac{x}{2}\right) \left(1 - \tan \left(\frac{x}{2}\right) \tan \left(\frac{h}{2}\right)\right)}{1 - \tan \left(\frac{x}{2}\right) \tan \left(\frac{h}{2}\right)}\right)$

$\frac{d}{\mathrm{dx}} \tan \left(\frac{x}{2}\right) = {\lim}_{h \to 0} \frac{1}{h} \left(\frac{\cancel{\tan \left(\frac{x}{2}\right)} + \tan \left(\frac{h}{2}\right) - \cancel{\tan \left(\frac{x}{2}\right)} + {\tan}^{2} \left(\frac{x}{2}\right) \tan \left(\frac{h}{2}\right)}{1 - \tan \left(\frac{x}{2}\right) \tan \left(\frac{h}{2}\right)}\right)$

$\frac{d}{\mathrm{dx}} \tan \left(\frac{x}{2}\right) = {\lim}_{h \to 0} \left(\tan \frac{\frac{h}{2}}{h}\right) \left(\frac{1 + {\tan}^{2} \left(\frac{x}{2}\right)}{1 - \tan \left(\frac{x}{2}\right) \tan \left(\frac{h}{2}\right)}\right)$

As:

${\lim}_{h \to 0} \tan \left(\frac{h}{2}\right) = 0$

${\lim}_{h \to 0} \frac{\tan \left(\frac{h}{2}\right)}{h} = \frac{1}{2} {\lim}_{h \to 0} \frac{\tan \left(\frac{h}{2}\right)}{\frac{h}{2}} = \frac{1}{2}$

$\frac{d}{\mathrm{dx}} \tan \left(\frac{x}{2}\right) = \frac{1}{2} \left(1 + {\tan}^{2} \left(\frac{x}{2}\right)\right) = \frac{1}{2} \left(1 + {\sin}^{2} \frac{\frac{x}{2}}{\cos} ^ 2 \left(\frac{x}{2}\right)\right) = \frac{1}{2} \frac{{\cos}^{2} \left(\frac{x}{2}\right) + {\sin}^{2} \left(\frac{x}{2}\right)}{\cos} ^ 2 \left(\frac{x}{2}\right) = \frac{1}{2 {\cos}^{2} \left(\frac{x}{2}\right)} = \frac{1}{2} {\sec}^{2} \left(\frac{x}{2}\right)$