# How do you find the derivative of csc (t/2)?

Mar 30, 2016

$- \frac{1}{2} \csc \left(\frac{t}{2}\right) \cot \left(\frac{t}{2}\right)$

#### Explanation:

Note that $\csc \left(\frac{t}{2}\right) = \frac{1}{\sin} \left(\frac{t}{2}\right)$

Using chain rule:
Let $u = \sin \left(\frac{t}{2}\right)$ thus the original function is $\frac{1}{u}$ and $\frac{\mathrm{du}}{\mathrm{dt}} = \frac{1}{2} \cos \left(\frac{t}{2}\right)$
$\frac{d}{\mathrm{du}} \frac{1}{u} = - \frac{1}{u} ^ 2 = - \frac{1}{{\sin}^{2} \left(\frac{t}{2}\right)}$
Thus $\frac{d}{\mathrm{dt}} \csc \left(\frac{t}{2}\right) = \frac{d}{\mathrm{du}} \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dt}} = - \frac{1}{{\sin}^{2} \left(\frac{t}{2}\right)} \cdot \frac{1}{2} \cos \left(\frac{t}{2}\right) = - \cos \frac{\frac{t}{2}}{2 {\sin}^{2} \left(\frac{t}{2}\right)} = - \frac{1}{2} \csc \left(\frac{t}{2}\right) \cot \left(\frac{t}{2}\right)$

Or by quotient rule:

$\left(\frac{f}{g}\right) ' = \frac{f ' g - g ' f}{g} ^ 2$
Let $f = 1$, $g = \sin \left(\frac{t}{2}\right)$
$f ' = 0$,$g ' = \frac{1}{2} \cos \left(\frac{t}{2}\right)$
$\left(\frac{f}{g}\right) ' = \frac{f ' g - g ' f}{{g}^{2}} = \frac{- \frac{1}{2} \cos \left(\frac{t}{2}\right)}{\sin} ^ 2 \left(\frac{t}{2}\right) = - \frac{1}{2} \csc \left(\frac{t}{2}\right) \cot \left(\frac{t}{2}\right)$