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Question Number 108766 by bemath last updated on 19/Aug/20

$$\frac{\mathcal{B}e\mathcal{M}ath}{★}$$$$\left(1\right)findtheequationofthetangentlineto$$$$thegraphoftheequation{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)=\frac{\pi }{2}$$$$atgivenpoint(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$$$$\left(2\right)Iff\left(x\right)=\underset{t\to x}{\lim }\frac{\sec t-\sec x}{t-x},findthevalueof$$$$f{}^{\prime }\left(\frac{\pi }{4}\right)$$$$\left(3\right)\underset{x\to 1}{\lim }\frac{{\tan }^{-1}\left(x\right)-\frac{\pi }{4}}{x-1}$$

Answered by Dwaipayan Shikari last updated on 19/Aug/20

$$3)\underset{x\to 1}{\lim }\frac{{tan}^{-1}\frac{x-1}{1+x}}{x-1}=\frac{x-1}{x-1}.\frac{1}{1+x}=\frac{1}{2}$$

Answered by john santu last updated on 19/Aug/20

$$\frac{⊸JS⊸}{\mathrm{♡}}$$$$\left(1\right)\frac{d}{dx}[{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(y\right)]=0$$$$\Rightarrow \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}}{y}^{\prime }\left(x\right)=0$$$$gradient\Rightarrow y^{\prime }\left(\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{1-\frac{1}{2}}}{\sqrt{1-\frac{1}{2}}}=1$$$$\therefore theequationoftangentline$$$$y=1.(x-\frac{\sqrt{2}}{2})+\frac{\sqrt{2}}{2};y=x$$$$\left(2\right)f\left(x\right)=\underset{t\to x}{\lim }\frac{\sec t-\sec x}{t-x}$$$$\Rightarrow f\left(x\right)=\underset{t\to x}{\lim }\frac{\cos x-\cos t}{\cos t.\cos x(t-x)}$$$$\Rightarrow f\left(x\right)=\underset{t\to x}{\lim }\frac{1}{\cos t.\cos x}.\underset{t\to x}{\lim }\frac{-2\sin \left(\frac{x+t}{2}\right)\sin \left(\frac{x-t}{2}\right)}{t-x}$$$$\Rightarrow f\left(x\right)=\frac{\sin x}{\cos {}^{2}x}\to f{}^{\prime }\left(x\right)=\frac{\cos {}^{3}x+2\sin {}^{2}x\cos x}{\cos {}^{4}x}$$$$\Rightarrow f{}^{\prime }\left(x\right)=\frac{\cos {}^{2}x+2\sin {}^{2}x}{\cos {}^{3}x}$$$$thereforef{}^{\prime }\left(\frac{\pi }{4}\right)=\frac{\frac{1}{2}+2.\frac{1}{2}}{\frac{1}{2\sqrt{2}}}=\frac{3}{2}.2\sqrt{2}=3\sqrt{2}$$$$\left(3\right)\underset{x\to 1}{\lim }\frac{{\tan }^{-1}\left(x\right)-\frac{\pi }{4}}{x-1}=\underset{x\to 1}{\lim }\left[\frac{1}{1+x^2}\right]=\frac{1}{2}$$

Commented by bemath last updated on 19/Aug/20

$$nice...$$

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

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

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

$$2)\mathrm{f}\left(\mathrm{x}\right)={\lim }_{\mathrm{t}\to \mathrm{x}}\frac{\frac{1}{\mathrm{sint}}-\frac{1}{\mathrm{sinx}}}{\mathrm{t}-\mathrm{x}}\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{sinx}}\right)=-\frac{\mathrm{cosx}}{{\sin }^2\mathrm{x}}\Rightarrow$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{x}\right)=-\frac{-{\mathrm{sinxsin}}^2\mathrm{x}-2\mathrm{sinx}\mathrm{cosx}\mathrm{cosx}}{{\sin }^4\mathrm{x}}$$$$=\frac{{\sin }^2\mathrm{x}+{2\cos }^2\mathrm{x}}{{\sin }^3\mathrm{x}}=\frac{1+{\cos }^2\mathrm{x}}{{\sin }^3\mathrm{x}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\frac{\pi }{4}\right)=\frac{1+{\left(\frac{1}{\sqrt{2}}\right)}^2}{{\left(\frac{1}{\sqrt{2}}\right)}^3}=\frac{3}{2\left(\frac{1}{2\sqrt{2}}\right)}=3\sqrt{2}$$

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

$$3){\lim }_{\mathrm{x}\to 1}\frac{\mathrm{arctanx}-\frac{\pi }{4}}{\mathrm{x}-1}={\lim }_{\mathrm{x}\to 1}\frac{\arctan \left(\mathrm{x}\right)-\arctan \left(1\right)}{\mathrm{x}-1}$$$$={\arctan }^{{}^{\prime }}\left(1\right)=\frac{1}{2}$$

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