If y= [2 ^ -1 1-x 1+x ], find dy dx . - Vidyadip

Question

If $\text{y}=\sin\Big[2\tan^{-1}\Big\{\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big\}\Big],$ find $\frac{\text{dy}}{\text{dx}}.$

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Answer

Here, $\text{y}=\sin\Big[2\tan^{-1}\Big[\sqrt{\frac{1-\text{x}}{1+\text{x}}}\Big]\Big]$
Put, $\text{x}=\cos2\theta, \text{So},$
$\text{y}=\sin\Big[2\tan^{-1}\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}\Big]$
$=\sin\Big[2\tan^{-1}\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}\Big]$
$=\sin\big[2\tan^{-1}\sqrt{\tan^2\theta}\big]$
$=\sin\big[2\tan^{-1}(\tan\theta)\big]$
$=\sin{2\theta}$
$=\sin\Big[2\times\frac{1}{2}\cos^{-1}\text{x}\Big]\ \big[\text{Since, x}=\cos2\theta\big]$
$=\sin\big(\sin^{-1}\sqrt{1-\text{x}^2}\big)$
$\text{y}=\sqrt{1-\text{x}^2}$
Differentiating with respect to x using chain rule,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}\frac{\text{d}}{\text{dx}}(1-\text{x}^2)$
$\frac{\text{dy}}{\text{dx}}=\frac{1}{2\sqrt{1-\text{x}^2}}(-2\text{x})$
$\frac{\text{dy}}{\text{dx}}=\frac{-\text{x}}{\sqrt{1-\text{x}^2}}$

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