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Question Number 118969 by benjo_mathlover last updated on 21/Oct/20

$$Givenf\left(x\right)=\frac{\sin x+\cos x}{\sin x-\cos x}$$$$findthevalueof$$$$f{}^{″}\left(x\right)+f{}^{\prime }\left(x\right)+1.$$

Answered by TANMAY PANACEA last updated on 21/Oct/20

$$f\left(x\right)=\frac{tanx+1}{tanx-1}=-tan(\frac{\pi }{4}+x)$$$$f^{{}^{\prime }}\left(x\right)=-{sec}^2(\frac{\pi }{4}+x)$$$$f^{{}^{″}}\left(x\right)=-2{sec}^2(\frac{\pi }{4}+x)tan(\frac{\pi }{4}+x)$$$$1+f^{{}^{\prime }}\left(x\right)+f^{{}^{″}}\left(x\right)=1-2{sec}^2(\frac{\pi }{4}+x)tan(\frac{\pi }{4}+x)-{sec}^2(\frac{\pi }{4}+x)$$$$$$

Answered by bramlexs22 last updated on 21/Oct/20

$$f{}^{\prime }\left(x\right)=\frac{(\cos x-\sin x)(\sin x-\cos x)-(\sin x+\cos x)(\cos x+\sin x)}{1-\sin 2x}$$$$=\frac{\sin 2x-1-(1+\sin 2x)}{1-\sin 2x}=\frac{-2}{1-\sin 2x}$$$$f{}^{″}\left(x\right)=\frac{0-(-2)(-2\cos 2x)}{{(1-\sin 2x)}^2}=\frac{-4\cos 2x}{{(1-\sin 2x)}^2}$$$$$$$$f{}^{″}\left(x\right)+f{}^{\prime }\left(x\right)+1=\frac{-4\cos 2x-2(1-\sin 2x)+{(1-\sin 2x)}^2}{{(1-\sin 2x)}^2}$$$$=\frac{-4\cos 2x+2\sin 2x-2+1-2\sin 2x+\sin {}^{2}2x}{{(1-\sin 2x)}^2}$$$$=\frac{\sin {}^{2}2x-4\cos 2x-1}{{(1-\sin 2x)}^2}=\frac{1-\cos {}^{2}2x-4\cos 2x-1}{{(1-\sin 2x)}^2}$$$$=\frac{-\cos 2x(\cos 2x+4)}{{(1-\sin 2x)}^2}$$

Answered by MJS_new last updated on 21/Oct/20

$$\sin x=\frac{\tan x}{\sqrt{1+{\tan }^{2}x}}\wedge \cos x=\frac{1}{\sqrt{1+{\tan }^{2}x}}$$$$\mathrm{let}t=\tan x\Rightarrow t^{\prime }=1+t^2$$$$\Rightarrow$$$$f\left(x\right)=\frac{t+1}{t-1}$$$$f^{\prime }\left(x\right)=-\frac{2t^{\prime }}{{(t-1)}^2}=-\frac{2(t^2+1)}{{(t-1)}^2}$$$$f^{″}\left(x\right)=\frac{4t^{\prime }(t+1)}{{(t-1)}^3}=\frac{4(t+1)(t^2+1)}{{(t-1)}^3}$$$$1+f^{\prime }\left(x\right)+f^{″}\left(x\right)=\frac{(t+1)(3t^2+5)}{{(t-1)}^3}=$$$$=-\frac{(1+\tan x)(5+{3\tan }^{2}x)}{{(1-\tan x)}^3}=$$$$=\frac{(\sin x+\cos x)(3+{2\cos }^{2}x)}{{(\sin x-\cos x)}^3}$$

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