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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!