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If $f(x)=x \tan ^{-1} x$, then $f(1)=$
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Verified Answer
The correct answer is:
$\frac{1}{2}+\frac{\pi}{4}$
$f(x)=x \tan ^{-1} x$
Differentiating w.r.t $x$, we get
$f^{\prime}(x)=x \frac{1}{1+x^2}+\tan ^{-1} x$
Now put $x=1$, then $f^{\prime}(1)=\frac{1}{2}+\tan ^{-1}(1)=\frac{\pi}{4}+\frac{1}{2}$.
Differentiating w.r.t $x$, we get
$f^{\prime}(x)=x \frac{1}{1+x^2}+\tan ^{-1} x$
Now put $x=1$, then $f^{\prime}(1)=\frac{1}{2}+\tan ^{-1}(1)=\frac{\pi}{4}+\frac{1}{2}$.
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