Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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

$${whats}\:{the}\:{a}\:{nswer}\:{right} \\ $$

Commented by Mrsof last updated on 05/Jul/21

???????

$$??????? \\ $$

Commented by Mrsof last updated on 05/Jul/21

how can solve this

$${how}\:{can}\:{solve}\:{this} \\ $$

Answered by Ar Brandon last updated on 05/Jul/21

tanh^(−1) (y)=lnx ⇒y=tanh(lnx) ⇒y′=(1/x)∙sech^2 (lnx)=(1/x)((2/((e^(lnx) +e^(−lnx) ))))^2 =(4/x)∙(x+(1/x))^(−2) =(4/x)(x^2 +2+(1/x^2 ))^(−1) =(4/x)(((x^4 +2x^2 +1)/x^2 ))^(−1) =(4/x)∙(x^2 /((x^2 +1)^2 )) =((4x)/((x^2 +1)^2 ))

$$\mathrm{tanh}^{−\mathrm{1}} \left(\mathrm{y}\right)=\mathrm{lnx}\:\Rightarrow\mathrm{y}=\mathrm{tanh}\left(\mathrm{lnx}\right) \\ $$$$\Rightarrow\mathrm{y}'=\frac{\mathrm{1}}{\mathrm{x}}\centerdot\mathrm{sech}^{\mathrm{2}} \left(\mathrm{lnx}\right)=\frac{\mathrm{1}}{\mathrm{x}}\left(\frac{\mathrm{2}}{\left(\mathrm{e}^{\mathrm{lnx}} +\mathrm{e}^{−\mathrm{lnx}} \right)}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{x}}\centerdot\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{−\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{x}}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)^{−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{x}}\left(\frac{\mathrm{x}^{\mathrm{4}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)^{−\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{x}}\centerdot\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4x}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Answered by Mathspace last updated on 05/Jul/21

tsnh^(−1) y=logx ⇒y=tanh(logx) =((sh(logx))/(ch(logx)))=((e^(logx) −e^(−logx) )/(e^(logx) +e^(−logx) )) =((x−(1/x))/(x+(1/x)))=((x^2 −1)/(x^2 +1))=1−(2/((x^2 +1))) ⇒ y^′ (x)=−2×((−2x)/((x^2 +1)^2 ))=((4x)/((x^2 +1)^2 ))

$${tsnh}^{−\mathrm{1}} {y}={logx}\:\Rightarrow{y}={tanh}\left({logx}\right) \\ $$$$=\frac{{sh}\left({logx}\right)}{{ch}\left({logx}\right)}=\frac{{e}^{{logx}} −{e}^{−{logx}} }{{e}^{{logx}} +{e}^{−{logx}} } \\ $$$$=\frac{{x}−\frac{\mathrm{1}}{{x}}}{{x}+\frac{\mathrm{1}}{{x}}}=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}=\mathrm{1}−\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$${y}^{'} \left({x}\right)=−\mathrm{2}×\frac{−\mathrm{2}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{4}{x}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com