Question

What is the derivative of coth(lnx)? I've been going over this one for a long time and can't get my answer to match the back of the book which is -4x / (1-x^2)^2

Answer 1

would you agree that \[\cot h (\ln x) \implies \frac{\cosh (\ln x)}{\sinh (\ln x)}\]

Answer 2

I would agree

Answer 3

would it be best to convert to that format before taking the derivative? \[\frac{ x-x ^{-1} }{ x + x ^{-1}{} }\]

Answer 4

no need. just use quotient rule here

Answer 5

\[f'(x) = \frac{ \frac{ \sinh ^{2}lnx }{ x } - \frac{ \cosh ^{2}lnx }{ x } }{ \sinh ^{2}lnx }\]how does that look so far?

Answer 6

so you have \[\frac{\frac 1x [\sinh^2 (\ln x) - \cosh^2 (\ln x)]}{\sinh^2 (\ln x)}\]
right?

Answer 7

yes