Question

find dy/dx of the given attachment​

Answer 1

$\boxed{\textbf{\large{Step-by-step explanation:}}}$

$y = log( \sqrt{( \frac{1 + cosx}{1 - cosx} } )$

$y = log( \frac{ \sqrt{(1 + cosx)} }{ \sqrt{(1 - cosx)} } )$

$y = log \sqrt{(1 + cosx)} - log \sqrt{(1 - cosx)}$

Differentiation wrt x

$\frac{dy}{dx} = \frac{d}{dx} (log \sqrt{(1 + cosx)} - log \sqrt{(1 -cosx) }$

$= \frac{1}{2 \sqrt{(1 + cosx)} } \times (0 - sinx) - \frac{1}{2 \sqrt{(1 - cosx)} } \times (0 + sinx)$

$= \frac{ - sinx}{2 \sqrt{(1 + cosx)} } - \frac{sinx}{2 \sqrt{(1 - cosx)} }$

$= \frac{ - sin(2 \sqrt{(1 - cosx)} - sinx(2 \sqrt{(1 + cosx)} }{4 \sqrt{(1 + cosx)(1 - cosx)} }$

$= \frac{ - 2sinx( \sqrt{(1 - cosx)} + \sqrt{(1 + cosx)} }{4( \sqrt{ {(1)}^{2} - {cos}^{2} x}) }$

$= \frac{ - sin( \sqrt{(1 - cosx)} + \sqrt{(1 + cosx)} )}{2(sinx)}$

$= -\frac{1}{2} \times (( \sqrt{1 - cosx) } + \sqrt{1 + cosx} )$

$\frac{dy}{dx}= - \frac{1}{2} \times (( \sqrt{1 - cosx) } + \sqrt{1 + cosx} )$