Question Number 222908 by Nicholas666 last updated on 10/Jul/25
Answered by Ak090 last updated on 10/Jul/25
$${d}/{dx}\:{ln}\left({cosh}\:{x}\right)\:=\:{tanh}\:{x} \\ $$$${d}/{dx}\:^{{p}} \sqrt{{x}}\:\mathrm{using}\:\mathrm{the}\:\mathrm{formula}\:{d}/{dx}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$=\:\mathrm{1}/{f}'\left({f}^{−\mathrm{1}} \left({x}\right)\right)\:=\:\mathrm{1}/{px}^{\left({p}−\mathrm{1}\right)/{p}} \\ $$$$\mathrm{so}\:{d}/{dx}\:^{{p}} \sqrt{{cosh}\:{x}}\:=\:{sinh}\:{x}\:/\:{pcosh}^{\left({p}−\mathrm{1}\right)/{p}} \\ $$$$\mathrm{same}\:\mathrm{for}\:\mathrm{the}\:\mathrm{other}\:\mathrm{root} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{denote}\:\mathrm{the}\:\mathrm{new}\:\mathrm{radicals}\:\mathrm{by}\:\mathrm{their}\:\mathrm{upper}\:\mathrm{case}\:\mathrm{letters} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{becomes}\:{tanh}\:{x}\:/\left({sinh}\:{x}\:/\:{P}\:−\:{sinh}\:{x}\:/\:{Q}\right) \\ $$$$=\:{tanh}\:{x}\:/\left(\:{Qsinh}\:{x}\:−\:{Psinh}\:{x}\:\right)/\left({PQ}\right) \\ $$$$={tanh}\:{x}\:/\:{sinh}\:{x}\:\left({Q}−{P}\right)/{PQ} \\ $$$$={PQ}/\left({cosh}\:{x}\right)/{Q}−{P} \\ $$$$={PQsech}\:{x}\:/\:{Q}−{P}\:\:\:\:\mathrm{return}\:\mathrm{to}\:\mathrm{the}\:\mathrm{original}\:\mathrm{form} \\ $$$${pqcosh}^{\left({p}−\mathrm{1}\right)/{p}+\left({q}−\mathrm{1}\right)/{q}} {x}\:{sech}\:{x}\:/\left({qcosh}^{\left({q}−\mathrm{1}\right)/{q}} −{pcosh}^{\left({p}−\mathrm{1}\right)/{p}} \right) \\ $$$$\mathrm{to}\:\mathrm{delete}\:\mathrm{sech}\:\mathrm{x}\:\mathrm{multiply}\:\mathrm{the}\:\mathrm{numerator}\:\mathrm{and}\:\mathrm{denominator}\:\mathrm{by}\:\mathrm{cosh}\:\mathrm{x} \\ $$$$\mathrm{it}\:\mathrm{becomes} \\ $$$${pqcosh}^{\left({p}−\mathrm{1}\right)/{p}+\left({q}−\mathrm{1}\right)/{q}} {x}/\left({qcosh}\:{x}−{pcosh}\:{x}\right) \\ $$$${substitute}\:{x}\:=\:\mathrm{0}\:\mathrm{it}\:\mathrm{becomes} \\ $$$${pq}/\left({q}−{p}\right)\:\:\:\:\:\left({Q}.{E}.{D}\right) \\ $$