Question #72ad1

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Answer 1 · 2017-09-29T15:07:44

$y'=3/4*csch(2x)$

1) I used $(Lnu)'$ = $(u')/u$ rule.

2) I used $(sechu)^2/tanhu=2csch(2u)$ rule.

$y=3/4*ln[tanh(x/2)]$

$y'=3/4*1/2*[sech(x/2)]^2/tanh(x/2)$

$y'=3/8*[sech(x/2)]^2/tanh(x/2)$

$y'=3/8*1/(coshx*sinhx)$

$y'=3/8*1/(1/2*sinh(2x))$

$y'=3/4*csch(2x)$