Question Number 145489 by Mrsof last updated on 05/Jul/21
Commented by Mrsof last updated on 05/Jul/21
$${whats}\:{the}\:{a}\:{nswer}\:{right} \\ $$
Commented by Mrsof last updated on 05/Jul/21
$$??????? \\ $$
Commented by Mrsof last updated on 05/Jul/21
$${how}\:{can}\:{solve}\:{this} \\ $$
Answered by Ar Brandon last updated on 05/Jul/21
$$\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{y}\right)=\mathrm{lnx}\:\Rightarrow\mathrm{y}=\mathrm{tanh}\left(\mathrm{lnx}\right) \\ $$$$\Rightarrow\mathrm{y}'=\frac{\mathrm{1}}{\mathrm{x}}\centerdot\mathrm{sech}^{\mathrm{2}} \left(\mathrm{lnx}\right)=\frac{\mathrm{1}}{\mathrm{x}}\left(\frac{\mathrm{2}}{\left(\mathrm{e}^{\mathrm{lnx}} +\mathrm{e}^{−\mathrm{lnx}} \right)}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{x}}\centerdot\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{−\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{x}}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)^{−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{x}}\left(\frac{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)^{−\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{x}}\centerdot\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4x}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Answered by Mathspace last updated on 05/Jul/21
$${tsnh}^{−\mathrm{1}} {y}={logx}\:\Rightarrow{y}={tanh}\left({logx}\right) \\ $$$$=\frac{{sh}\left({logx}\right)}{{ch}\left({logx}\right)}=\frac{{e}^{{logx}} −{e}^{−{logx}} }{{e}^{{logx}} +{e}^{−{logx}} } \\ $$$$=\frac{{x}−\frac{\mathrm{1}}{{x}}}{{x}+\frac{\mathrm{1}}{{x}}}=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}=\mathrm{1}−\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$${y}^{'} \left({x}\right)=−\mathrm{2}×\frac{−\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{4}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$