How do you differentiate #f(x) = (cosx)/(sinx)# using the quotient rule?

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Answer 1 · 2017-01-03T15:23:05

#f'(x) = -csc^2x#

Let #f(x) = (g(x))/(h(x))#, then #g(x) = cosx# and #h(x) = sinx#. Then #g'(x) = -sinx# and #h'(x) = cosx#.

#f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2#

#f'(x) = (-sinx(sinx) - cosx(cosx))/(sinx)^2#

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

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

#f'(x) = -1/sin^2x#

#f'(x) = -csc^2x#

Hopefully this helps!