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Question Number 38762 by Raj Singh last updated on 29/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18

Commented by abdo.msup.com last updated on 29/Jun/18

$$wehavef\left(x\right)=arctan(cosx+sinx)$$$$=arctan\left(\sqrt{2}cos(x-\frac{\pi }{4})\right)\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{-\sqrt{2}sin(x-\frac{\pi }{4})}{1+2{cos}^2(x-\frac{\pi }{4})}=\sqrt{2}sin(\frac{\pi }{4}-x)$$$$wehave0⩽x⩽\frac{\pi }{4}\Rightarrow -\frac{\pi }{4}⩽-x⩽0\Rightarrow$$$$0⩽\frac{\pi }{4}-x⩽\frac{\pi }{4}\Rightarrow \sqrt{2}sin(\frac{\pi }{4}-x)⩾0\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)⩾0\Rightarrow fisincreasingon[0,\frac{\pi }{4}].$$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jun/18

$$y={tan}^{-1}(sinx+cosx)$$$$\frac{dy}{dx}=\frac{cosx-sinx}{1+{(sinx+cosx)}^2}$$$$when\frac{\mathrm{\Pi }}{4}>x>0valueofcosx>sinx$$$$so\frac{dy}{dx}>0hencey=f\left(x\right)increasingfunction$$$$intheinterval(0,\frac{\mathrm{\Pi }}{4})$$$$iamattachinggraphforbetterunderstanding$$$$fo$$$$$$

Answered by MJS last updated on 29/Jun/18

$$\sin x+\cos x=\sqrt{2}\sin (x+\frac{\pi }{4})$$$$f\left(x\right)=arctan\left(\sqrt{2}\sin (x+\frac{\pi }{4})\right)$$$$\frac{d}{dx}\left[\arctan f\left(x\right)\right]=\frac{f^{\prime }\left(x\right)}{f{\left(x\right)}^2+1}=$$$$=\frac{\sqrt{2}\cos (x+\frac{\pi }{4})}{{2\sin }^2(x+\frac{\pi }{4})+1}$$$$$$$$({2\sin }^2(x+\frac{\pi }{4})+1)>0\mathrm{\forall }x\in \mathbb{R}$$$$\sqrt{2}\cos (x+\frac{\pi }{4})=0\Rightarrow x=z\pi -\frac{3}{4}\pi ;z\in \mathbb{Z}\Rightarrow$$$$\Rightarrow \sqrt{2}\cos (x+\frac{\pi }{4})>0\mathrm{\forall }x\in ]-\frac{3}{4}\pi ;\frac{\pi }{4}[\Rightarrow$$$$\Rightarrow f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{for}x\in ]-\frac{3}{4}\pi ;\frac{\pi }{4}[\Rightarrow$$$$\Rightarrow f\left(x\right)\mathrm{is}\mathrm{increasing}\mathrm{for}x\in ]0;\frac{\pi }{4}[$$$$$$

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