Pls solve these qstns i hav to submit it day after tommorrow

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{y}={\tan }^{-1}\left(\sqrt{{\mathrm{x}}^2-1}\right) \\ \mathrm{Differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{y}}_1=\frac{1}{1+{\left(\sqrt{{\mathrm{x}}^2-1}\right)}^2}.\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{{\mathrm{x}}^2-1}\right) \\ =\frac{1}{1+{\mathrm{x}}^2-1}.\frac{1}{2\sqrt{{\mathrm{x}}^2-1}}.\left(2\mathrm{x}\right)=\frac{\mathrm{x}}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^2-1}} \\ \Rightarrow {\mathrm{y}}_1=\frac{1}{\mathrm{x}\sqrt{{\mathrm{x}}^2-1}}.....\left(\mathrm{i}\right) \\ \mathrm{Again} \mathrm{differentiate} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x} \\ \Rightarrow {\mathrm{y}}_2=-\frac{1}{{\left(\mathrm{x}\sqrt{{\mathrm{x}}^2-1}\right)}^2}.\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}\sqrt{{\mathrm{x}}^2-1}\right) \\ =-\frac{1}{{\left(\mathrm{x}\sqrt{{\mathrm{x}}^2-1}\right)}^2}\left\{\mathrm{x}.\frac{1}{2\sqrt{{\mathrm{x}}^2-1}}.2\mathrm{x}+\sqrt{{\mathrm{x}}^2-1}\right\} \\ =-\frac{1}{{\mathrm{x}}^2\left({\mathrm{x}}^2-1\right)}\left\{\frac{{\mathrm{x}}^2}{\sqrt{{\mathrm{x}}^2-1}}+\sqrt{{\mathrm{x}}^2-1}\right\} \\ =-\frac{1}{{\mathrm{x}}^2\left({\mathrm{x}}^2-1\right)}\left\{\frac{{\mathrm{x}}^2+{\mathrm{x}}^2-1}{\sqrt{{\mathrm{x}}^2-1}}\right\} \\ \Rightarrow {\mathrm{y}}_2=-\frac{2{\mathrm{x}}^2-1}{{\mathrm{x}}^2\left({\mathrm{x}}^2-1\right)\sqrt{{\mathrm{x}}^2-1}}...\left(\mathrm{ii}\right) \\ \mathrm{L}.\mathrm{H}.\mathrm{S}. \\ =\left(2{\mathrm{x}}^2-1\right){\mathrm{y}}_1+\mathrm{x}\left({\mathrm{x}}^2-1\right){\mathrm{y}}_2 \\ =\left(2{\mathrm{x}}^2-1\right)\left(\frac{1}{\mathrm{x}\sqrt{{\mathrm{x}}^2-1}}\right)+\mathrm{x}\left({\mathrm{x}}^2-1\right)\left(-\frac{2{\mathrm{x}}^2-1}{{\mathrm{x}}^2\left({\mathrm{x}}^2-1\right)\sqrt{{\mathrm{x}}^2-1}}\right) \left[\mathrm{Use} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right)\right] \\ =\frac{2{\mathrm{x}}^2-1}{\mathrm{x}\sqrt{{\mathrm{x}}^2-1}}-\frac{2{\mathrm{x}}^2-1}{\mathrm{x}\sqrt{{\mathrm{x}}^2-1}} \\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}. \left(\mathrm{Hence} \mathrm{Proved}\right) \\ \mathrm{Please} \mathrm{don}\text{'}\mathrm{t} \mathrm{ask} \mathrm{all} \mathrm{questions} \mathrm{in} \mathrm{same} \mathrm{thread}. \end{gathered}$$

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