What is the derivative of #f(x) = cot(x)#?

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Answer 1 · 2015-10-28T21:01:50

Derivative of #cot(x)# is equal to #-csc^2(x)#.

We know that #cot(x) = 1/tan(x)# so #f'(x) = 1/tan(x)dx#

We can use the quotient rule to solve for the derivative. The quotient rule states:

#d(g(x)/(h(x))) = ((g'(x)h(x) - g(x)h'(x))/g(x)^2)dx#

in our case,

#g(x) = 1#
#h(x) = tan(x)#
#g'(x) = 0#
#h'(x) = sec^2(x)#

Let's plug these values back into the quotient rule:

#(0 * tan(x) - 1 * sec^2(x))/tan^2(x) = -sec^2(x)/tan^2(x) = - csc^2(x)#