Question:
If $\mathrm{y}=\tan ^{-1}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$, find $\frac{\mathrm{dy}}{\mathrm{dx}}$.
Solution:
$y=\tan ^{-1}\left(\frac{1-x}{1+x}\right)$
Using the Chain Rule of Differentiation,
$\frac{d y}{d x}=\frac{1}{1+\left(\frac{1-x}{1+x}\right)^{2}} \cdot \frac{(1+x) \cdot(1-x)^{\prime}-(1+x)^{\prime} \cdot(1-x)}{(1+x)^{2}}$
$=\frac{(1+x)^{2}}{(1+x)^{2}+(1-x)^{2}} \cdot \frac{(1+x)(-1)-(1)(1-x)}{(1+x)^{2}}$
$=-\frac{2}{(1+x)^{2}+(1-x)^{2}}$
$=-\frac{1}{1+x^{2}}$ (Ans)
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