Question

If $f(t)=\frac{1-t}{1+t}$, then the value of $f^{\prime }(1/t)$ is

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

Answer 1

Answer: (A)

$f^{\prime }(t)=\frac{d}{dt}\left[\frac{1-t}{1+t}\right]=\frac{(1+t)(-1)-(1-t)\times (1)}{(1+t{)}^2}$
$=\frac{-1-t-1+t}{(1+t{)}^2}=\frac{-2}{(1+t{)}^2}$
$f^{\prime }[1/t]=\frac{-2}{{\left(1+\frac{1}{t}\right)}^2}=\frac{-2t^2}{(t+1{)}^2}$