Question:
Show that $f(x)=\tan ^{-1} x-x$ is a decreasing function on $R$ ?
Solution:
we have,
$f(x)=\tan ^{-1} x-x$
$f^{\prime}(x)=\frac{1}{1+x^{2}}-1$
$=-\frac{x^{2}}{1+x^{2}}$
Now,
$\mathrm{X} \in \mathrm{R}$
$\Rightarrow x^{2}>0$ and $1+x^{2}>0$
$\Rightarrow \frac{x^{2}}{1+x^{2}}>0$
$\Rightarrow-\frac{x^{2}}{1+x^{2}}<0$
$\Rightarrow f^{\prime}(x)<0$
Hence, $f(x)$ is an decreasing function for $R$
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