WAEC - Further Mathematics (2017 - No. 21)

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

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

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

\(\frac{-1}{\sqrt{1-x}}\)

\(\frac{1}{\sqrt{1-x}}\)

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

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

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

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