y=xx+1y = \frac{x}{x+1}y=x+1x
Using quotient rule because the function is of the form u(x)v(x)\frac{u(x)}{v(x)}v(x)u(x)
dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}dxdy=v2vdxdu−udxdv
dydx=(x+1)⋅1−x⋅1(x+1)2\frac{dy}{dx} = \frac{(x+1)\cdot 1 - x\cdot 1}{(x+1)^2}dxdy=(x+1)2(x+1)⋅1−x⋅1
= 1(x+1)2\frac{1}{(x+1)^2}(x+1)21
