How do you find the derivative of ##cotx##?

##dy/dx = -csc^2x##

##y = cotx##

##y = 1/tanx##

##y = 1/(sinx/cosx)##

##y = cosx/sinx##

Letting ##y= (g(x))/(h(x))##, we have that ##g(x) = cosx## and ##h(x) = sinx##.

##y' = (g'(x) xx h(x) - g(x) xx h'(x))/(h(x))^2##

##y' = (-sinx xx sinx - (cosx xx cosx))/(sinx)^2##

##y' = (-sin^2x - cos^2x)/(sinx)^2##

##y' = (-(sin^2x + cos^2x))/sin^2x##

##y' = -1/sin^2x##

##y' = -csc^2x##

Hopefully this helps!