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प्रश्न
If \[y = \sin^{- 1} \left( 6x\sqrt{1 - 9 x^2} \right), - \frac{1}{3\sqrt{2}} < x < \frac{1}{3\sqrt{2}}\] \[\frac{dy}{dx} \] ?
उत्तर
We have, \[y = \sin^{- 1} \left( 6x\sqrt{1 - 9 x^2} \right), - \frac{1}{3\sqrt{2}} < x < \frac{1}{3\sqrt{2}}\]
\[So, \frac{dy}{dx} = \frac{d}{dx}\left[ \sin^{- 1} \left( 6x\sqrt{1 - 9 x^2} \right) \right]\]
\[ = \frac{d}{dx}\left[ \sin^{- 1} \left( 6x\sqrt{1 - 9 x^2} \right) \right]\]
\[ = \frac{1}{\sqrt{1 - \left( 6x\sqrt{1 - 9 x^2} \right)^2}} \times \frac{d}{dx}\left( 6x\sqrt{1 - 9 x^2} \right)\]
\[ = \frac{1}{\sqrt{1 - \left[ 36 x^2 \left( 1 - 9 x^2 \right) \right]}} \times \left( 6x\frac{d}{dx}\sqrt{1 - 9 x^2} + \sqrt{1 - 9 x^2}\frac{d}{dx}\left( 6x \right) \right)\]
\[ = \frac{1}{\sqrt{1 - 36 x^2 - 324 x^4}} \times \left( 6x \times \frac{1}{2\sqrt{1 - 9 x^2}}\frac{d}{dx}\left( 1 - 9 x^2 \right) + \sqrt{1 - 9 x^2}\left( 6 \right) \right)\]
\[ = \frac{1}{\sqrt{1 - 36 x^2 - 324 x^4}} \times \left( 6x \times \frac{1}{2\sqrt{1 - 9 x^2}} \times \left( - 18x \right) + 6\sqrt{1 - 9 x^2} \right)\]
\[ = \frac{1}{\sqrt{1 - 36 x^2 - 324 x^4}} \times \left( \frac{- 54 x^2}{\sqrt{1 - 9 x^2}} + 6\sqrt{1 - 9 x^2} \right)\]
\[ = \frac{1}{\sqrt{1 - 36 x^2 - 324 x^4}} \times \left( \frac{- 54 x^2 + 6\left( 1 - 9 x^2 \right)}{\sqrt{1 - 9 x^2}} \right)\]
\[ = \frac{- 54 x^2 + 6 - 54 x^2}{\sqrt{1 - 9 x^2}\sqrt{1 - 36 x^2 - 324 x^4}}\]
\[ = \frac{6 - 108 x^2}{\sqrt{1 - 9 x^2}\sqrt{1 - 36 x^2 - 324 x^4}}\]
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