How can you prove that $d/dx(cothx) = -csch^2x$ using the definition $cothx=coshx/sinhx$ ?

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Answer 1 · 2018-07-11T05:53:31

Please see the proof below

We need

$(coshx)'=sinhx$

$cothx=coshx/sinhx$

$cosh^2x-sinh^2x=1$

$(u/v)'=(u'v-uv')/(v^2)$

$u=coshx$ , $=>$ , $u'=sinhx$

$v=sinhx$ , $=>$ , $v'=coshx$

Therefore,

$(cothx)'=(sinh^2x-cosh^2x)/(sinh^2x)=-1/sinh^2x=csch^2x$

How can you prove that d/dx(cothx) = -csch^2xd/dx(cothx) = -csch^2x using the definition cothx=coshx/sinhxcothx=coshx/sinhx?