Question

If $f(t)=\frac{1-t}{1+t}$, then the value of $f'(1 / t)$ is

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

Answer 1

Answer: (A)

$f'(t) =\frac{d}{d t}\left[\frac{1-t}{1+t}\right]=\frac{(1+t)(-1)-(1-t) \times(1)}{(1+t)^{2}}$
$=\frac{-1-t-1+t}{(1+t)^{2}}=\frac{-2}{(1+t)^{2}}$
$f'[1 / t] =\frac{-2}{\left(1+\frac{1}{t}\right)^{2}}=\frac{-2 t^{2}}{(t+1)^{2}}$