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 $cosxsinx=(sin2x)/2$ .

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

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