How do I find the derivative of #y = arc cot(x) + (cot x)^-1#?

I'll assume that you don't know what the derivative of #"arccot"(x)# is, so that we can use implicit differentiation to find #(dy)/dx#.

First, start by rewriting your function as

#y = "arccot"(x) + 1/cotx" "#, which is equivalent to

#y = "arccot"(x) + tan(x)#

Rerrange to get #arccos(x)# alone on the right-hand side of the equation

#y - tan(x) = "arccot"(x)#

This is equivalent to

#cot(y -tan(x)) = x " "color(blue)((1))#

Differentiate both sides with respect to #x# to get

#d/dxcot(y - tan(x)) = d/dx(x)#

#-csc^2(y - tan(x)) * [(dy)/dx - sec^2(x)] = 1#

#-csc^2(y - tan(x)) * (dy)/dx + csc^2(y - tan(x)) * sec^2(x) = 1#

Rearrange to get #(dy)/dx# alone on the left-hand side of the equation

#-csc^2(y - tan(x)) * (dy)/dx = 1 - csc^2(y - tan(x)) * sec^2(x)#

#(dy)/dx = (1 - csc^2(y - tan(x)) * sec^2(x))/(-csc^2(y - tan(x)))#

#(dy)/dx = -1/csc^2(y - tan(x)) + (color(red)(cancel(color(black)(-csc^2(y - tan(x))))) * sec^2(x))/color(red)(cancel(color(black)(-csc^2(y - tan(x))))#

#(dy)/dx = sec^2x - 1/csc^2(y - tan(x))#

Use the trigonometric identity

#color(blue)(csc^2x = 1 + cot^2x)#

to write

#(dy)/dx = sec^2x - 1/(1 + cot^2(y - tan(x)))#

You know from equation #color(blue)((1))# that

#cot(y - tan(x)) = x#

which means that the derivative will be

#(dy)/dx = color(green)(sec^2x - 1/(1 + x^2))#