Search any question & find its solution
Question:
Answered & Verified by Expert
If \(f(t)=\frac{1-t}{1+t}\), then \(f^{\prime}(1 / t)\) is equal to
Options:
Solution:
1056 Upvotes
Verified Answer
The correct answer is:
\(\frac{-2 t^2}{(t+1)^2}\)
\(\begin{aligned}
& f^{\prime}(t)=\frac{d}{d t}\left[\frac{1-t}{1+t}\right]=\frac{(1+t)(-1)-(1-t) \times(1)}{(1+t)^2} \\
& =\frac{-1-t-1+t}{(1+t)^2}=\frac{-2}{(1+t)^2}
\end{aligned}\)
\(f^{\prime}[1 / t]=\frac{-2}{\left(1+\frac{1}{t}\right)^2}=\frac{-2 t^2}{(t+1)^2}\)
& 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}\)
\(f^{\prime}[1 / t]=\frac{-2}{\left(1+\frac{1}{t}\right)^2}=\frac{-2 t^2}{(t+1)^2}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.