Question

If \(\rm y = \dfrac{x+1}{x-1}\), then what is \(\rm \dfrac{dy}{dx}\) equal to?
1. \(\rm \dfrac{-2}{x-1}\)
2. \(\rm \dfrac{-2}{(x-1)^2}\)
3. \(\rm \dfrac{2}{(x-1)^2}\)
4. \(\rm \dfrac{2}{(x-1)}\)
5. None of these

Answer 1

Correct Answer - Option 2 : \(\rm \dfrac{-2}{(x-1)^2}\)

Concept:

\(\rm \frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x) f^{'}(x)-f(x) g^{\prime}(x)}{[g(x)]^{2}}\)

Calculation:

Here, \(\rm y = \dfrac{x+1}{x-1}\)

\(\rm \dfrac{dy}{dx}\)

\(\begin{array}{l} =\frac{(x-1)(1)-(x+1)(1)}{(x-1)^{2}} \\ =\frac{x-1-x-1}{(x-1)^{2}} \end{array}\)

\(=\rm \dfrac{-2}{(x-1)^2}\)

Hence, option (2) is correct.