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Question Number 145489 by Mrsof last updated on 05/Jul/21

Commented by Mrsof last updated on 05/Jul/21

$$whatstheanswerright$$

Commented by Mrsof last updated on 05/Jul/21

$$???????$$

Commented by Mrsof last updated on 05/Jul/21

$$howcansolvethis$$

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

$${\tanh }^{-1}\left(\mathrm{y}\right)=\mathrm{lnx}\Rightarrow \mathrm{y}=\tanh \left(\mathrm{lnx}\right)$$$$\Rightarrow {\mathrm{y}}^{\prime }=\frac{1}{\mathrm{x}}\cdot {\mathrm{sech}}^2\left(\mathrm{lnx}\right)=\frac{1}{\mathrm{x}}{\left(\frac{2}{({\mathrm{e}}^{\mathrm{lnx}}+{\mathrm{e}}^{-\mathrm{lnx}})}\right)}^2$$$$=\frac{4}{\mathrm{x}}\cdot {(\mathrm{x}+\frac{1}{\mathrm{x}})}^{-2}=\frac{4}{\mathrm{x}}{({\mathrm{x}}^2+2+\frac{1}{{\mathrm{x}}^2})}^{-1}$$$$=\frac{4}{\mathrm{x}}{\left(\frac{{\mathrm{x}}^4+{2\mathrm{x}}^2+1}{{\mathrm{x}}^2}\right)}^{-1}=\frac{4}{\mathrm{x}}\cdot \frac{{\mathrm{x}}^2}{{({\mathrm{x}}^2+1)}^2}$$$$=\frac{4\mathrm{x}}{{({\mathrm{x}}^2+1)}^2}$$

Answered by Mathspace last updated on 05/Jul/21

$${tsnh}^{-1}y=logx\Rightarrow y=tanh\left(logx\right)$$$$=\frac{sh\left(logx\right)}{ch\left(logx\right)}=\frac{e^{logx}-e^{-logx}}{e^{logx}+e^{-logx}}$$$$=\frac{x-\frac{1}{x}}{x+\frac{1}{x}}=\frac{x^2-1}{x^2+1}=1-\frac{2}{(x^2+1)}\Rightarrow$$$$y^{{}^{\prime }}\left(x\right)=-2\times \frac{-2x}{{(x^2+1)}^2}=\frac{4x}{{(x^2+1)}^2}$$

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