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

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Answer 1 · 2016-02-10T16:53:29

#f'(x)=-2/(sin2x)#

We will have to use the chain rule:

#d/dx(ln(x))=1/x" "=>" "d/dx(ln(g(x)))=1/(g(x))*g'(x)#

Thus,

#f'(x)=1/cotx*d/dx(cotx)#

Since the derivative of #cotx# is #-csc^2x#,

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

We could simplify this in terms of sine and cosine:

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

We could further simplify this by recognizing that #2cosxsinx=sin2x#. so #cosxsinx=(sin2x)/2#.

#f'(x)=-1/((sin2x)/2)=-2/(sin2x)#