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The function $y=\tan ^{-1} x-x \quad$
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is always decreasing
Let $y=\tan ^{-1} x-x$
On differentiating w.r.t $x$, we get $\frac{d y}{d x}=\frac{1}{1+x^{2}}-1=\frac{1-1-x^{2}}{1+x^{2}}=\frac{-x^{2}}{1+x^{2}}$
$\Rightarrow \frac{d y}{d x} < 0, \forall x \in R$
Hence, function is always decreasing.
On differentiating w.r.t $x$, we get $\frac{d y}{d x}=\frac{1}{1+x^{2}}-1=\frac{1-1-x^{2}}{1+x^{2}}=\frac{-x^{2}}{1+x^{2}}$
$\Rightarrow \frac{d y}{d x} < 0, \forall x \in R$
Hence, function is always decreasing.
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