Find dy dx if y = ^ -1 [ 6x - 41 - 4x^ 2 5 ] - Vidyadip

Question

Find $\frac{\text{dy}}{\text{dx}} \text{if y = }\sin^{-1}\bigg[\frac{6x - 4\sqrt{1 - 4x}^{2}}{5}\bigg]$

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Answer

$\text{Let 2x} = \sin\theta$
$\therefore\text{y} = \sin^{-1}\bigg(\frac{\text{6x} - \sqrt{1 - 4\text{x}^{2}}}{5}\bigg)$
$= \sin^{-1}\bigg(\frac{3}{5}\sin\theta- \frac{4}{5}\cos\theta\bigg)$
$= \sin^{-1}(\cos \alpha \sin \theta - \sin \alpha \cos \theta) $ $[\cos\alpha = \frac{3}{5}; \sin \alpha = \frac{4}{5}]$

$= \sin^{-1}(\sin\theta - \alpha)$

$= \theta - \alpha$

$= \sin^{-1}\text{(2x)} - \alpha$

$\Rightarrow\frac{\text{dy}}{\text{dx}} = \frac{2}{\sqrt{1 - 4x}^{2}}$

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