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Question Number 39949 by zesearho last updated on 13/Jul/18

$$f\left(x\right)=2arctan\left(\sqrt{\frac{1-x}{x}}\right)$$$$f^{\prime }\left(x\right)=?$$

Commented by abdo mathsup 649 cc last updated on 13/Jul/18

$$directcalculus$$$$f^{{}^{\prime }}\left(x\right)=2\frac{{\left(\sqrt{\frac{1-x}{x}}\right)}^{{}^{\prime }.}}{1+\frac{1-x}{x}}but$$$${\sqrt{\frac{1-x}{x}})}^{{}^{\prime }}={\left(\sqrt{\frac{1}{x}-1}\right)}^{{}^{\prime }}=\frac{-\frac{1}{x^2}}{2\sqrt{\frac{1-x}{x}}}=-\frac{1}{2x^2\sqrt{\frac{1-x}{x}}}$$$$f^{,}\left(x\right)=-2x\frac{1}{2x^2\sqrt{\frac{1-x}{x}}}=-\frac{1}{x\sqrt{\frac{1-x}{x}}}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Jul/18

$$y=2{tan}^{-}\left(\sqrt{\frac{1-x}{x}}\right)letx={sin}^2\theta$$$${cos}^2\theta =1-x$$$$x={sin}^2\theta so\frac{dx}{d\theta }=2sin\theta cos\theta$$$$y=2{tan}^{-1}\left(\frac{cos\theta }{sin\theta }\right)$$$$y=2{tan}^{-1}\left(cot\theta \right)$$$$y=2{tan}^{-1}\left\{tan(\frac{\mathrm{\Pi }}{2}-\theta )\right\}=\mathrm{\Pi }-2\theta$$$$\frac{dy}{d\theta }=-2$$$$$$$$$$$$$$$$\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=\frac{-2}{2sin\theta cos\theta }=\frac{-1}{sin\theta cos\theta }$$$$\frac{-1}{\sqrt{x}}\times \frac{1}{\sqrt{1-x}}$$

Answered by ajfour last updated on 13/Jul/18

$$y={2\tan }^{-1}\sqrt{\frac{1-x}{x}}$$$$\Rightarrow \tan {}^2\frac{y}{2}=\frac{1}{x}-1$$$$2\tan \frac{y}{2}(1+\tan {}^2\frac{y}{2})\times \frac{1}{2}\frac{dy}{dx}=-\frac{1}{x^2}$$$$\Rightarrow \frac{dy}{dx}=\frac{-1}{x^2\sqrt{\frac{1-x}{x}}(1+\frac{1}{x}-1)}$$$$f{}^{\prime }\left(x\right)=\frac{-1}{x}\sqrt{\frac{x}{1-x}}.$$

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