Math, asked by sweetysiri92, 1 year ago

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

Answers

Answered by TPS
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

TPS: thanks!!
Answered by kvnmurty
1
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
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