Derivative of $\tan{\left(\frac{x}{2} \right)}$

Find $\frac{d}{dx} \left(\tan{\left(\frac{x}{2} \right)}\right)$.

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 $\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$:

The derivative of the tangent is $\frac{d}{du} \left(\tan{\left(u \right)}\right) = \sec^{2}{\left(u \right)}$:

Return to the old variable:

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 $f{\left(x \right)} = x$:

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dx} \left(x\right) = 1$:

Simplify:

Thus, $\frac{d}{dx} \left(\tan{\left(\frac{x}{2} \right)}\right) = \frac{1}{\cos{\left(x \right)} + 1}$.

$\frac{d}{dx} \left(\tan{\left(\frac{x}{2} \right)}\right) = \frac{1}{\cos{\left(x \right)} + 1}$A