Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 66866 by rajesh4661kumar@gmail.com last updated on 20/Aug/19

Commented by mathmax by abdo last updated on 20/Aug/19

$$firstx+\sqrt{1+x^2}>0forallxsoy=ln(x+\sqrt{1+x^2})=argsh\left(x\right)\Rightarrow$$$$y^{{}^{\prime }}\left(x\right)=\frac{1}{\sqrt{1+x^2}}={(1+x^2)}^{-\frac{1}{2}}\Rightarrow y^{{}^{″}}\left(x\right)=-\frac{1}{2}\left(2x\right){(1+x^2)}^{-\frac{3}{2}}$$$$=\frac{-x}{{(1+x^2)}^{\frac{3}{2}}}\Rightarrow (x^2+1)y^{{}^{″}}+{xy}^{\prime }=(x^2+1)\frac{-x}{{(1+x^2)}^{\frac{3}{2}}}+\frac{x}{\sqrt{1+x^2}}$$$$=\frac{-x}{\sqrt{1+x^2}}+\frac{x}{\sqrt{1+x^2}}=0sotheQmustbe(x^2+1)y^{{}^{″}}+{xy}^{{}^{\prime }}=0$$$$$$

Answered by Kunal12588 last updated on 20/Aug/19

$$y=log(x+\sqrt{x^2+1})$$$$\Rightarrow y^{\prime }=\frac{1}{x+\sqrt{x^2+1}}\times (1+\frac{2x}{2\sqrt{x^2+1}})$$$$\Rightarrow y^{\prime }=\frac{1}{x+\sqrt{x^2+1}}\times \left(\frac{x+\sqrt{x^2+1}}{\sqrt{x^2+1}}\right)$$$$\Rightarrow y^{\prime }=\frac{1}{\sqrt{x^2+1}}$$$$\Rightarrow y^{\prime }\sqrt{x^2+1}=1$$$$\Rightarrow \sqrt{x^2+1}\times y^{″}+y^{\prime }\times \frac{2x}{2\sqrt{x^2+1}}=0$$$$\Rightarrow (x^2+1)y^{″}+{xy}^{\prime }=0$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com