Question

If f(x) = square root of secx-1/ secx +1 , find f'(x) also find f'(x) (pi/3)=?​

Answer 1

$Given \: \: Question \: \: Is \: \: \\ \\ f(x) = \sqrt{ \frac{ \sec(x) - 1 }{ \sec(x) + 1 } } \\ \\Answer \\ \\ \: Replace \: \: \: \: \: \sec(x) \: \: \: by \: \: \: \frac{1}{ \cos(x) } \: \: \: \: we \: have \\ \\ f(x) = \sqrt{ \frac{ \frac{1}{ \cos(x) } - 1 }{ \frac{1}{ \cos(x) } + 1 } } \\ \\ \\ f(x) = \sqrt{ \frac{1 - \cos(x) }{1 + \cos(x) } } \\ \\ \\ 1 - \cos(x) = 2 \sin {}^{2} ( \frac{x}{2} ) \: \: \: \\ \\ and \: \: \\ \\ 1 + \cos( \frac{x}{2} ) = 2 \cos {}^{2} ( \frac{x}{2} ) \\ \\ \\ f(x) = \sqrt{ \frac{2 \sin {}^{2} ( \frac{x}{2} ) }{2 \cos {}^{2} ( \frac{x}{2} ) } } \\ \\ \\ f(x) = \sqrt{ \tan {}^{2} ( \frac{x}{2} ) } \\ \\ \\ f(x) = \tan( \frac{x}{2} ) \\ \\ Differentiate \: \: both \: \: sides \: \: with \: \: \\ respect \: \: to \: x \: \: we \: \: \: have. \\ \\ f {}^{1} (x) = \frac{1}{2} \sec {}^{2} ( \frac{x}{2} ) \\ \\ therefore \: \: Differentiation \: \: of \: \: \\ \sqrt{ \frac{ \sec(x) - 1}{ \sec(x) + 1 } } \: \: \: is \: \: \: \frac{1}{2} \sec {}^{2} ( \frac{x}{2} ) \\ \\ \\ for \: \: \: f {}^{1} ( \frac{\pi}{3} ) \: \: \: put \: \: \: x = \frac{\pi}{3} \: \: \: we \: \: have \\ \\ \\ f {}^{1} ( \frac{\pi}{3} ) = \frac{1}{2} \sec {}^{2} ( \frac{\pi}{6} ) \\ \\ \\ f {}^{1} ( \frac{\pi}{3} ) = \frac{1}{2} ( \frac{2}{ \sqrt{3} } ) {}^{2} \\ \\ \\ f {}^{1} ( \frac{\pi}{3} ) = \frac{2}{3}$