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Derivative of $$$\tan{\left(\frac{x}{2} \right)}$$$
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Find $$$\frac{d}{dx} \left(\tan{\left(\frac{x}{2} \right)}\right)$$$.
Solution
The function $$$\tan{\left(\frac{x}{2} \right)}$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \tan{\left(u \right)}$$$ and $$$g{\left(x \right)} = \frac{x}{2}$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\tan{\left(\frac{x}{2} \right)}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right) \frac{d}{dx} \left(\frac{x}{2}\right)\right)}$$The derivative of the tangent is $$$\frac{d}{du} \left(\tan{\left(u \right)}\right) = \sec^{2}{\left(u \right)}$$$:
$${\color{red}\left(\frac{d}{du} \left(\tan{\left(u \right)}\right)\right)} \frac{d}{dx} \left(\frac{x}{2}\right) = {\color{red}\left(\sec^{2}{\left(u \right)}\right)} \frac{d}{dx} \left(\frac{x}{2}\right)$$Return to the old variable:
$$\sec^{2}{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right) = \sec^{2}{\left({\color{red}\left(\frac{x}{2}\right)} \right)} \frac{d}{dx} \left(\frac{x}{2}\right)$$Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = \frac{1}{2}$$$ and $$$f{\left(x \right)} = x$$$:
$$\sec^{2}{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(\frac{x}{2}\right)\right)} = \sec^{2}{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{\frac{d}{dx} \left(x\right)}{2}\right)}$$Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 1$$$, in other words, $$$\frac{d}{dx} \left(x\right) = 1$$$:
$$\frac{\sec^{2}{\left(\frac{x}{2} \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{2} = \frac{\sec^{2}{\left(\frac{x}{2} \right)} {\color{red}\left(1\right)}}{2}$$Simplify:
$$\frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} = \frac{1}{\cos{\left(x \right)} + 1}$$Thus, $$$\frac{d}{dx} \left(\tan{\left(\frac{x}{2} \right)}\right) = \frac{1}{\cos{\left(x \right)} + 1}$$$.
Answer
$$$\frac{d}{dx} \left(\tan{\left(\frac{x}{2} \right)}\right) = \frac{1}{\cos{\left(x \right)} + 1}$$$A