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Question Number 97353 by john santu last updated on 07/Jun/20

Answered by abdomathmax last updated on 07/Jun/20

$$6)\mathrm{let}\mathrm{f}\left(\mathrm{x}\right)=\arcsin \left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)\mathrm{and}\mathrm{g}\left(\mathrm{x}\right)=2\arctan \sqrt{\mathrm{x}}-\frac{\pi }{2}$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{{\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)}^{{}^{\prime }}}{\sqrt{1-{\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)}^2}}=\frac{2}{{(\mathrm{x}+1)}^2\sqrt{\frac{{(\mathrm{x}+1)}^2-{(\mathrm{x}-1)}^2}{{(\mathrm{x}+1)}^2}}}$$$$=\frac{2}{(\mathrm{x}+1)\sqrt{{\mathrm{x}}^2+2\mathrm{x}+1-{\mathrm{x}}^2+2\mathrm{x}-1}}=\frac{2}{(\mathrm{x}+1)2\sqrt{\mathrm{x}}}$$$$=\frac{1}{(\mathrm{x}+1)\sqrt{\mathrm{x}}}$$$${\mathrm{g}}^{{}^{\prime }}\left(\mathrm{x}\right)=2\frac{\frac{1}{2\sqrt{\mathrm{x}}}}{\sqrt{1+\mathrm{x}}}=\frac{1}{\sqrt{\mathrm{x}}(\mathrm{x}+1)}\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)+\mathrm{C}$$$$\mathrm{C}=\mathrm{f}\left(0\right)-\mathrm{g}\left(0\right)=-\frac{\pi }{2}-(-\frac{\pi }{2})=0\Rightarrow$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{g}\left(\mathrm{x}\right)(\mathrm{with}\mathrm{x}⩾0)$$

Commented by john santu last updated on 08/Jun/20

$$\mathrm{thank}\mathrm{you}$$

Answered by abdomathmax last updated on 07/Jun/20

$$8){\lim }_{\mathrm{x}\to 1^{+}}(\frac{3}{\mathrm{lnx}}-\frac{2}{(\mathrm{x}-1)})$$$$={\lim }_{\mathrm{x}\to 1^{+}}\frac{3\mathrm{x}-3-2\mathrm{lnx}}{(\mathrm{x}-1)\mathrm{lnx}}$$$$={\lim }_{\mathrm{x}\to 1^{+}}\frac{3-\frac{2}{\mathrm{x}}}{\mathrm{lnx}+\frac{\mathrm{x}-1}{\mathrm{x}}}={\lim }_{\mathrm{x}\to 1^{+}}\frac{3\mathrm{x}-2}{\mathrm{xlnx}+\mathrm{x}-1}$$$$={\lim }_{\mathrm{x}\to 1^{+}}\frac{3}{\mathrm{lnx}+1+1}=\frac{3}{2}$$

Answered by abdomathmax last updated on 07/Jun/20

$${8)}^2{\lim }_{\mathrm{x}\to 1}\frac{\mathrm{arctanx}-\frac{\pi }{4}}{\mathrm{x}-1}$$$$={\lim }_{\mathrm{x}\to 1}\frac{\frac{1}{1+{\mathrm{x}}^2}}{1}=\frac{1}{2}$$

Answered by abdomathmax last updated on 07/Jun/20

$${8)}^3{\lim }_{\mathrm{x}\to 4^{+}}{\left(3(\mathrm{x}-4)\right)}^{\mathrm{x}-4}$$$$={\lim }_{\mathrm{x}\to 4^{+}}{\mathrm{e}}^{(\mathrm{x}-4)\ln \left(3(\mathrm{x}-4)\right)}$$$$={\lim }_{\mathrm{x}\to 4^{+}}{\mathrm{e}}^{(\mathrm{x}-4)\ln \left(3\right)+(\mathrm{x}-4)\ln (\mathrm{x}-4)}$$$$={\mathrm{e}}^0=1$$$$$$

Answered by abdomathmax last updated on 07/Jun/20

$$9)\mathrm{arcsinx}+\mathrm{arcosy}=\frac{\pi }{2}\Rightarrow$$$$\mathrm{arcosy}=\frac{\pi }{2}-\mathrm{arcsinx}\Rightarrow$$$$\mathrm{y}=\cos (\frac{\pi }{2}-\mathrm{arcsinx})=\mathrm{x}\mathrm{equation}\mathrm{of}\mathrm{tangent}$$$$\mathrm{is}\mathrm{y}={\mathrm{f}}^{{}^{\prime }}\left(\frac{\sqrt{2}}{2}\right)(\mathrm{x}-\frac{\sqrt{2}}{2})+\mathrm{f}\left(\frac{\sqrt{2}}{2}\right)$$$$=1(\mathrm{x}-\frac{\sqrt{2}}{2})+\frac{\sqrt{2}}{2}\Rightarrow \mathrm{y}=\mathrm{x}$$

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