# Where are the critical points of csc x?

The critical points of $\csc x$ will occur at those values of $x$ for which $\frac{d}{\mathrm{dx}} \csc x = 0$. By using the quotient rule, which states that for $f \left(x\right) = g \frac{x}{h \left(x\right)} , \frac{d}{\mathrm{dx}} f \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$, one can obtain the derivative of $\csc x$ (though examining a table of derivatives for trigonometric functions would also serve well in this regard).
Since $\csc x = \frac{1}{\sin} x$, by use of the derivative of
$f \left(x\right) = \frac{1}{\sin} x , g \left(x\right) = 1 , h \left(x\right) = \sin x , g ' \left(x\right) = \frac{d}{\mathrm{dx}} 1 = 0 , h ' \left(x\right) = \frac{d}{\mathrm{dx}} \sin x = \cos x$ , we obtain:
$\frac{d}{\mathrm{dx}} \csc x = \frac{\left(0\right) \left(\sin x\right) - \left(1\right) \left(\cos x\right)}{\sin x} ^ 2 = - \cos \frac{x}{\sin x} ^ 2$
We know that $\cos \frac{x}{\sin} \left(x\right) = \cot \left(x\right)$ and $\frac{1}{\sin} \left(x\right) = \csc \left(x\right)$, so this yields...
$\frac{d}{\mathrm{dx}} \csc x = - \cot \left(x\right) \csc \left(x\right)$.
The critical points will occur where this derivative function is equal to 0. Looking at our function, specifically in its earlier form of $- \cos \frac{x}{\sin} ^ 2 \left(x\right)$, we realize that the function will only equal $0$ when $\cos \left(x\right) = 0$ (and, coincidentally, that the function will be undefined at every $x$ such that $\sin \left(x\right) = 0$. Assuming then that x is defined in radians, the critical points of the function $\csc x$ will occur at every $x = n \pi + \frac{\pi}{2}$, wherein $n$ is defined as any integer. Note that we could define the critical points as occurring at every $x = m \frac{\pi}{2}$ with $m$ being any odd integer.