If \(y = \frac{1+x}{1-x}\), find \(\frac{dy}{dx}\).

  • A \(\frac{2}{(1-x)^{2}}\)
  • B \(\frac{-2}{(1-x)^{2}}\)
  • C \(\frac{-1}{\sqrt{1-x}}\)
  • D \(\frac{1}{\sqrt{1-x}}\)
waec 2017furthermathematics

The correct answer is A. \(\frac{2}{(1-x)^{2}}\)

\(y = \frac{1+x}{1-x}\)

Using quotient rule, \(\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}\), we have

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

= \(\frac{2}{(1-x)^{2}}\).

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