How do you find the derivative of #cotx#?

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Answer 1 · 2016-11-14T04:43:21

#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!