সমাধান করঃ ${\sin }^{-1}x- {\cos }^{-1}x =\frac{\mathrm{\pi }}{6}$

$\mathrm{Let}, \mathrm{I}=\int 2^{\ln \mathrm{x}}\mathrm{dx}=2^{\ln \mathrm{x}}\int dx-\int \left({\frac{d}{dx}}^{\left(2^{\ln x}\right)}\int dx\right)dx$

$=2^{\ln x}.x-\int 2^{\ln \mathrm{x}}.ln2.\frac{1}{x}.x dx=x 2^{\ln \mathrm{x}}-ln2\int 2^{\ln \mathrm{x}}dx$

$=x 2^{\ln \mathrm{x}}-I.ln2$

$\therefore I+I ln2=x 2^{\ln \mathrm{x}}$

$\Rightarrow I\left(I+ln2\right)=x 2^{\ln \mathrm{x}}$

$\Rightarrow I=\frac{x 2^{\ln \mathrm{x}}}{1+ln2}+C$

$\underset{\mathrm{x}\to 0}{\lim }\frac{{\mathrm{e}}^{\mathrm{x}}-1}{\mathrm{x}}=\underset{\mathrm{x}\to 0}{\lim }\frac{\left(1+\frac{\mathrm{x}}{1!}+\frac{{\mathrm{x}}^2}{2!}+\frac{{\mathrm{x}}^3}{3!}+....\right)-1}{\mathrm{x}}$

$=\underset{\mathrm{x}\to 0}{\lim }\frac{x\left(1+\frac{x}{2!}+\frac{x^2}{3!}+...\right)}{x} =\underset{\mathrm{x}\to 0}{\lim }\frac{1+\frac{x}{2!}+\frac{x^2}{3!}+....}{1}$

$=1 \therefore L.H.S=R.H.S \left(Proved\right)$

$\underset{x\to 0}{\lim }\frac{3sinx-sin3x}{x^3}$

$=\underset{x\to 0}{\lim }\frac{4si{n}^{3}x}{x^3}$

$=\underset{x\to 0}{\lim }{\left(\frac{\sin x}{x}\right)}^3\times 4$

$=4\times 1=4 \left(Ans\right)$

$\frac{d}{dx}\left(x^{x^x}\right)=\frac{d}{dx}\left(e^{ln{x}^{x^x}}\right)=e^{{\ln }^{x^{x^x}}}\frac{d}{dx}\left(x^x \ln x\right)$

$=x^{x^x}\left\{x^x\frac{1}{x}+lnx\frac{d}{dx}\left(x^x\right)\right\}$

$=x^{x^x}\left\{x^x\frac{1}{x}+lnx.x^x\frac{d}{dx}\left(x lnx\right)\right\}$

$=x^{x^x}\left\{x^x\frac{1}{x}+lnx.x^x\left(x\frac{1}{x} lnx\right)\right\}$

$=x^{x^x} x^x\left\{\frac{1}{x}+lnx\left(1+lnx\right)\right\} \left(Ans\right)$

$\mathrm{y}={\left({\cos }^{-1}\mathrm{x}\right)}^2$

$\Rightarrow {\mathrm{y}}_1=2\left({\cos }^{-1}\mathrm{x}\right)\frac{-1}{\sqrt{1-{\mathrm{x}}^2}}$

$\Rightarrow {\mathrm{y}}_1^2\left(1-{\mathrm{x}}^2\right)=4{\left({\cos }^{-1}\mathrm{x}\right)}^2$

$\Rightarrow {\mathrm{y}}_1^2\left(1-{\mathrm{x}}^2\right)=4\mathrm{y}$

$\therefore {\mathrm{y}}_1^2\left(-2x\right)+\left(1-x^2\right)2{\mathrm{y}}_1{\mathrm{y}}_2=4{\mathrm{y}}_1$

$\Rightarrow \left(1-x^2\right){\mathrm{y}}_2-{\mathrm{xy}}_1=2 \left(\mathrm{Proved}\right)$