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.
