WAEC - Further Mathematics (2017 - No. 21)
If \(y = \frac{1+x}{1-x}\), find \(\frac{dy}{dx}\).
\(\frac{2}{(1-x)^{2}}\)
\(\frac{-2}{(1-x)^{2}}\)
\(\frac{-1}{\sqrt{1-x}}\)
\(\frac{1}{\sqrt{1-x}}\)
توضیح
\(y = \frac{1+x}{1-x}\)
Using quotient rule, \(\frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^{2}}\), we have
\(\frac{dy}{dx} = \frac{(1-x)(1) - (1+x)(-1)}{(1-x)^{2}} = \frac{(1 - x +1 +x)}{(1-x)^{2}}\)
= \(\frac{2}{(1-x)^{2}}\).