Question

If y= log tan(π/4+x/2),show that dy/dx=secx

Answer 1

$\textbf{Given:}$

$y=\log\,tan(\frac{\pi}{4}+\frac{x}{2})$

$\textbf{To find:}\;\dfrac{dy}{dx}$

$\text{We apply chain rule to find the derivtive the given function}$

$y=\log\,tan(\frac{\pi}{4}+\frac{x}{2})$

$\text{Differentiate with respect to x}$

$\dfrac{dy}{dx}=\dfrac{1}{tan(\frac{\pi}{4}+\frac{x}{2})}\,\dfrac{d(tan(\frac{\pi}{4}+\frac{x}{2}))}{dx}$

$\dfrac{dy}{dx}=\dfrac{1}{tan(\frac{\pi}{4}+\frac{x}{2})}\,sec^2(\frac{\pi}{4}+\frac{x}{2})\,\dfrac{d(\frac{x}{2})}{dx}$

$\dfrac{dy}{dx}=\dfrac{1}{tan(\frac{\pi}{4}+\frac{x}{2})}\,sec^2(\frac{\pi}{4}+\frac{x}{2})\,\dfrac{1}{2}$

$\dfrac{dy}{dx}=\dfrac{cos(\frac{\pi}{4}+\frac{x}{2})}{sin(\frac{\pi}{4}+\frac{x}{2})}\,\dfrac{1}{cos^2(\frac{\pi}{4}+\frac{x}{2})}\,\dfrac{1}{2}$

$\dfrac{dy}{dx}=\dfrac{1}{sin(\frac{\pi}{4}+\frac{x}{2})}\,\dfrac{1}{cos^2(\frac{\pi}{4}+\frac{x}{2})}\,\dfrac{1}{2}$

$\dfrac{dy}{dx}=\dfrac{1}{2\;sin(\frac{\pi}{4}+\frac{x}{2})\;cos(\frac{\pi}{4}+\frac{x}{2})}$

$\text{Using,}$

$\bf\;2\;sinA\;cosA=sin2A$

$\dfrac{dy}{dx}=\dfrac{1}{sin2(\frac{\pi}{4}+\frac{x}{2})}$

$\dfrac{dy}{dx}=\dfrac{1}{sin(\frac{2\pi}{4}+\frac{2x}{2})}$

$\dfrac{dy}{dx}=\dfrac{1}{sin(\frac{\pi}{2}+x)}$

$\dfrac{dy}{dx}=\dfrac{1}{cosx}$

$\implies\bf\dfrac{dy}{dx}=secx$

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