The critical points of csc x will occur at those values of x for which d/(dx) csc x = 0. By using the quotient rule, which states that for f(x) = g(x)/(h(x)), d/dx f(x) = (g'(x)h(x) - g(x)h'(x))/(h(x))^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 = 1/sin x, by use of the derivative of
f(x) = 1/sin x, g(x) = 1, h(x)=sin x, g'(x)=d/dx 1 = 0, h'(x) = d/dx sin x = cos x , we obtain:
d/dx csc x = ((0)(sin x) - (1)(cos x))/(sin x)^2 = -cos(x)/(sin x)^2
We know that cos(x)/sin(x) = cot(x) and 1/sin(x)=csc(x), so this yields...
d/dx cscx = -cot(x)csc(x).
The critical points will occur where this derivative function is equal to 0. Looking at our function, specifically in its earlier form of -cos(x)/sin^2(x), we realize that the function will only equal 0 when cos(x) = 0 (and, coincidentally, that the function will be undefined at every x such that sin(x) =0. Assuming then that x is defined in radians, the critical points of the function csc x will occur at every x = npi + pi/2, wherein n is defined as any integer. Note that we could define the critical points as occurring at every x = mpi/2 with m being any odd integer.
