50 POINTS !!!!! HomeWorker Helperer please1) Using the quotient rule, find [tex]y = \frac{cosx - sinx}{cosx + sinx}[/tex]2) Given that [tex]y = \frac{x-1}{x+1}[/tex] , prove that [tex](x+1)(\frac{d^{2}y }{dx^{2} }) + 2(\frac{dy}{dx}) = 0[/tex]

Answer:1)     - 2 / (cos x + sinx)^2.Step-by-step explanation:1) I assume you want the derivative of (cos x - sin x) / ( cos x + sin x).y'  =  (cosx + sinx)( -sinx - cosx) - (cosx - sinx)(-sinx + cosx)         ---------------------------------------------------------------------------                                  (cosx + sinx)^2=  -sinxcosx - cos^2x - sin^2x - sinxcosx - (-sinxcosx+cos^2x+sin^2x-sinxcosx        -------------------------------------------------------------------------------------------------------                                      (cosx + sinx)^2=     -2sinxcosx - 1  - ( -2sinxcosx   + 1 )   /     (cosx + sinx)^2=      -2sinxcosx - 1  - ( -2sinxcosx   + 1 )   /     (cosx + sinx)^2 =       - 2 / (cos x + sinx)^2.