Question
If $\text{y}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big),$, find $\frac{\text{dy}}{\text{dx}}.$
✓
Answer
Here,
$\text{y}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Differentiating it with respect to x using chain rule and quotinet rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\frac{1-\text{x}}{1+\text{x}}\big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
$=\frac{(1+\text{x})^2}{(1+\text{x}^2+2\text{x}+1+\text{x}^2-2\text{x})}\bigg[\frac{(1+\text{x})\frac{\text{d}}{\text{dx}}(1-\text{x})-(1-\text{x})\frac{\text{d}}{\text{dx}}(1+\text{x})}{(1+\text{x})^2}\bigg]$
$=\frac{(1+\text{x})^2}{2\text{x}^2+2}\Big[\frac{(1+\text{x})(-1)-(1-\text{x})(1)}{(1+\text{x})^2}\Big]$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\Big(\frac{-\text{x}-1-1+\text{x}}{(1+\text{x})^2}\Big)$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\times\frac{-2}{(1+\text{x})^2}$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\text{x}^2+1}$
$\text{y}=\tan^{-1}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
Differentiating it with respect to x using chain rule and quotinet rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\big(\frac{1-\text{x}}{1+\text{x}}\big)^2}\frac{\text{d}}{\text{dx}}\Big(\frac{1-\text{x}}{1+\text{x}}\Big)$
$=\frac{(1+\text{x})^2}{(1+\text{x}^2+2\text{x}+1+\text{x}^2-2\text{x})}\bigg[\frac{(1+\text{x})\frac{\text{d}}{\text{dx}}(1-\text{x})-(1-\text{x})\frac{\text{d}}{\text{dx}}(1+\text{x})}{(1+\text{x})^2}\bigg]$
$=\frac{(1+\text{x})^2}{2\text{x}^2+2}\Big[\frac{(1+\text{x})(-1)-(1-\text{x})(1)}{(1+\text{x})^2}\Big]$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\Big(\frac{-\text{x}-1-1+\text{x}}{(1+\text{x})^2}\Big)$
$=\frac{(1+\text{x})^2}{2(\text{x}^2+1)}\times\frac{-2}{(1+\text{x})^2}$
$\frac{\text{dy}}{\text{dx}}=-\frac{1}{\text{x}^2+1}$
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