Q56 / 125 · BITSAT 2009 · Mathematics

If \(f(t)=\frac{1-t}{1+t}\), then \(f^{\prime}(1 / t)\) is equal to

  • A
    \(\frac{1}{(1+t)^2}\)
  • B
    \(\frac{1}{(t-1)^2}\)
  • Correct answer
    C
    \(\frac{-2 t^2}{(t+1)^2}\)
  • D
    \(\frac{2}{(t-1)^2}\)

Solution

\(\begin{aligned} & f^{\prime}(t)=\frac{d}{d t}\left[\frac{1-t}{1+t}\right]=\frac{(1+t)(-1)-(1-t) \times(1)}{(1+t)^2} \\ & =\frac{-1-t-1+t}{(1+t)^2}=\frac{-2}{(1+t)^2} \end{aligned}\) \(f^{\prime}[1 / t]=\frac{-2}{\left(1+\frac{1}{t}\right)^2}=\frac{-2 t^2}{(t+1)^2}\)

Hence, the correct answer is option: 'C'.

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