What are the first and second derivatives of $g(x) = lnsinx$ ?

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Answer 1 · 2015-12-24T17:56:43

Using chain rule, we can rename $u=sinx$ and solve it following chain rule definition: $(dy)/(dx)=(dy)/(du)(du)/(dx)$

$(dy)/(dx)=(1/u)*cosx$

$(dy)/(dx)=cosx/sinx$

This is the same as $tan^(-1)x$ or $1/tanx$ , which is the same as $cotx$

$(dy^2)/(d^2x)=-csc^2x$

What are the first and second derivatives of g(x) = lnsinx g(x) = lnsinx?