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Question Number 44622 by mondodotto@gmail.com last updated on 02/Oct/18

$$\mathbf{if}\mathit{y}=\mathbf{ln}\left[\mathbf{tan}(\frac{\mathit{\pi }}{4}+\frac{\mathit{x}}{2})\right]\mathbf{show}\mathbf{that}$$$$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\mathbf{sec}\mathit{x}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

$$\frac{dy}{dx}=\frac{1}{tan(\frac{\pi }{4}+\frac{x}{2})}\times {sec}^2(\frac{\pi }{4}+\frac{x}{2})\times \frac{1}{2}$$$$\frac{dy}{dx}=\frac{1}{2}\times \frac{cos(\frac{\pi }{4}+\frac{x}{2})}{sin(\frac{\pi }{4}+\frac{x}{2})}\times \frac{1}{{cos}^2(\frac{\pi }{4}+\frac{x}{2})}$$$$\frac{dy}{dx}=\frac{1}{2sin(\frac{\pi }{4}+\frac{x}{2})cos(\frac{\pi }{4}+\frac{x}{2})}=\frac{1}{sin(\frac{\pi }{2}+x)}=\frac{1}{cosx}=secx$$

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