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If $f(t)=\int_{-t}^t \frac{d x}{1+x^2}$, then $f^{\prime}(1)$ is
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Given $f(t)=\int_t^t \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_{-t}^t=2 \tan ^{-1} t$
Differentiating with respect to $t, \quad f^{\prime}(t)=\frac{2}{1+t^2}$
$\Rightarrow \quad f^{\prime}(1)=\frac{2}{2}=1$
Differentiating with respect to $t, \quad f^{\prime}(t)=\frac{2}{1+t^2}$
$\Rightarrow \quad f^{\prime}(1)=\frac{2}{2}=1$
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