If f(t) = _ , - t ^ ,t dx 1 + x^2 , then f'(1) is - Vidyadip

MCQ

If $f(t) = \int_{\, - t}^{\,t} {\frac{{dx}}{{1 + {x^2}}},} $ then $f'(1)$ is

A
$Zero$
B
$2/3$
C
$ - \,1$
✓
$1$

✓

Answer

Correct option: D.

$1$

d
(d) Given $f(t) = \int_{ - t}^t {\frac{{dx}}{{1 + {x^2}}}} $

$ = [{\tan ^{ - 1}}x]_{ - t}^t$

$ = 2{\tan ^{ - 1}}t$

Differentiating with respect to $t$,

$f'(t) = \frac{2}{{1 + {t^2}}}$

==> $f'(1) = \frac{2}{2} = 1$.

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