Question

differentiation of Y=cosx-sinx/cosx+sinx with respect to x prove that dy/dx+y^2+1=0

Answer 1

Given that $y=\frac{\cos x-\sin x}{\cos x+\sin x}$.

Differentiating,

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

Now,

$LHS=\frac{dy}{dx}+y^2+1=-\frac{2}{(\cos x+\sin x)^2} +\frac{(\cos x-\sin x)^2}{(\cos x+\sin x)^2} +1\\ LHS=\frac{dy}{dx}+y^2+1=\frac{-2+(\cos x-\sin x)^2+(\cos x+\sin x)^2}{(\cos x+\sin x)^2} \\ LHS=\frac{dy}{dx}+y^2+1=\frac{-2+2}{(\cos x+\sin x)^2} \\ LHS=\frac{dy}{dx}+y^2+1=0=RHS$

the proof is complete.

Answer 2

y=cosx-sinx/cosx+sinx,. divide to cosx up and down ,then y=(cosx/cosx)-(sinx/cosx)/(cosx/cosx)+(sinx/cosx)=1-tanx/1+tanx=tanπ/4-tanx/1+tanx. {cosπ/4=1}. y=tan(π/4-x) =cotx.....Dy/DX=sec^2(π/4-x)×-1=-sec^2(π/4-x). Dy/DX=-[1+tan^2(π/4-x)]=-(1+y^2). Dy/DX+y^2+1=0