Question

d/dx((sinx+cosx)/(sinx-cosx))​

Answer 1

Step-by-step explanation:

Let

$y = \frac{ \sin(x) + \cos(x) }{ \sin(x ) - \cos(x) }$

$= > \frac{dy}{dx} = \frac{( \sin(x) - \cos(x) ). \frac{dy}{dx} ( \sin(x) + \cos(x)) - ( \sin(x) + \cos(x)). \frac{dy}{dx} ( \sin(x) - \cos(x)) }{ {( \sin(x) - \cos(x) )}^{2} }$

$\frac{dy}{dx} = \frac{ ( \sin(x) - \cos(x) )( \cos(x) - \sin(x)) - ( \sin(x) + \cos(x) )( \cos(x) + \sin(x) )}{ {( \sin(x) - \cos(x) ) }^{2} }$

$\frac{dy}{dx} = \frac{{( \sin(x) + \cos(x) ) }^{2} - {( \sin(x) - \cos(x) )}^{2} }{{( \sin(x) - \cos(x) )}^{2}}$

$\frac{dy}{dx} = \frac{4 \sin(x) \cos(x) }{ \sin^{2} (x) + \cos^{2} (x) - 2 \sin(x) \cos(x) }$

$= \frac{dy}{dx} = \frac{4 \sin(x) \cos(x) }{1 - 2 \sin(x) \cos(x) }$

Hope this will help you