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#