Question

evaluate the differential dy for the curve y = x+1/x-1 at x = 2,.where dx = 1/2

Answer 1

$y = \frac{x+1}{x-1} \\ \\ \frac{dy}{dx} = \frac{d}{dx} (\frac{x+1}{x-1}) \\ \\ \frac{dy}{dx} = \frac{(x-1)* \frac{d}{dx}(x+1) -(x+1)* \frac{d}{dx}(x-1)}{ (x-1)^{2} } \\ \\ \frac{dy}{dx} = \frac{(x-1)*1 -(x+1)* 1}{ (x-1)^{2} } \\ \\ \frac{dy}{dx} = \frac{(x-1) -(x+1)}{ (x-1)^{2} } \\ \\ \frac{dy}{dx} = \frac{-2}{ (x-1)^{2} }$

dx = 1/2, x = 2

$\frac{dy}{dx} = \frac{-2}{ (x-1)^{2} } \\ \\ {dy}= \frac{-2}{ (x-1)^{2} } {dx} \\ \\ dy=\frac{-2}{ (2-1)^{2} }* \frac{1}{2} \\ \\ dy= \frac{-2}{1}* \frac{1}{2} =-1$

Answer 2

$y = \frac{x+1}{x-1}=1+\frac{2}{x-1}\\\\\frac{dy}{dx}=\frac{-2}{(x-1)^2}\\\\dy=\frac{-2}{(2-1)^2}(\frac{1}{2})=-1,$    On substituting x = 2 and dx = 1/2,  we get