You can differentiate this function by using the quotient rule, which tells you that the derivative of afunction expressed as the quotient of two other functions
#f(x) = g(x)/h(x)#
can be found by using
#color(blue)(d/dx(f(x)) = (g^'(x) * h(x) - g(x) * h^'(x))/[g(x)]^2)#, with #h(x)!=0#.
You can simplify this expression by using the trigonometric identity
#cot(x) = 1/tan(x) = cos(x)/sin(x)#
This means that you can write
#f(x) = cosx/sinx * 1/sinx = cosx/sin^2x#
This function's derivative will thus be
#d/dx(f(x)) = ([d/dx(cosx)] * sin^2x - cosx * d/dx(sin^2x))/(sin^2x)^2#
You can use the power and chain rules to find #d/dx(sin^2x)#.
More specifically, you can write #sin^2x = u^2#, with #u = sinx#. This will get you
#d/dx(u^2) = d/du(u^2) * d/dx(sinx)#
#d/dx(u^2) = 2u * cosx#
#d/dx(sin^2x) = 2sinx * cosx#
Plug this back into your target derivative to get
#f^' = (-sinx * sin^2x - cosx * 2 * sinx * cosx)/sin^4x#
#f^' = (-sin^3x - 2sinx cos^2x)/sin^4x#
You can simplify this further by
#f^' = - (color(red)(cancel(color(black)(sinx)))(sin^2x + 2cos^2x))/sin^(color(red)(cancel(color(black)(4))) color(blue)(3))x#
#f^' = -(overbrace(sin^2x + cos^2x)^(color(red)("=1")) + cos^2x)/sin^3x#
#f^' = color(green)(-(1 + cos^2x)/sin^3x)#
