find f'(x) by applying this formula:
$\displaystyle f'(x_1) = \lim_{x\to x_1} \frac {f(x) - f(x_1)}{x - x_1}$
$\displaystyle \csc x; x_1 = \frac {1}{2} \pi$
I'm going to have to assume we know two limits:
$\displaystyle \lim_{h \to 0} \frac{1 - cos(h)}{h} = 0$
$\displaystyle \lim_{h \to 0} \frac{sin(h)}{h} = 1$
(I'm going to use h instead of $\displaystyle x_1$ to save me a bit of coding.)
$\displaystyle f'(x) = \lim_{h \to 0} \frac{ csc(x+h) - csc(x) }{h}$
$\displaystyle f'(x) = \lim_{h \to 0} \frac{ \frac{1}{sin(x+h)} - \frac{1}{sin(x)} }{h}$
$\displaystyle f'(x) = \lim_{h \to 0} \frac{sin(x) - sin(x + h)}{hsin(x)sin(x+h)}$
Now expand the sin(x + h) = sin(x)cos(h) + sin(h)cos(x) in the numerator:
$\displaystyle f'(x) = \lim_{h \to 0} \frac{sin(x) - sin(x)cos(h) - sin(h)cos(x)}{hsin(x)sin(x+h)}$
$\displaystyle f'(x) = \lim_{h \to 0} \frac{sin(x) - sin(x)cos(h)}{hsin(x)sin(x+h)} - \lim_{h \to 0} \frac{sin(h)cos(x)}{hsin(x)sin(x+h)}$
$\displaystyle f'(x) = \lim_{h \to 0} \frac{sin(x)[1 - cos(h)]}{hsin(x)sin(x+h)} - \lim_{h \to 0} \frac{sin(h)}{h} \cdot \lim_{h \to 0}\frac{cos(x)}{sin(x)sin(x+h)}$
$\displaystyle f'(x) = \lim_{h \to 0} \frac{1 - cos(h)}{h} \cdot \lim_{h \to 0} \frac{1}{sin(x+h)} - \lim_{h \to 0} \frac{sin(h)}{h} \cdot \lim_{h \to 0} \frac{cos(x)}{sin(x)sin(x+h)}$
You can now see the limits I mentioned above. So:
$\displaystyle f'(x) = 0 \cdot \lim_{h \to 0} \frac{1}{sin(x+h)} - 1 \cdot \lim_{h \to 0} \frac{cos(x)}{sin(x)sin(x+h)}$
$\displaystyle f'(x) = -\lim_{h \to 0} \frac{cos(x)}{sin(x)sin(x+h)}$
$\displaystyle f'(x) = -\frac{cos(x)}{sin^2(x)}$
-Dan