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# Solution for If Y = Sin − 1 ( 6 X √ 1 − 9 X 2 ) , − 1 3 √ 2 < X < 1 3 √ 2 D Y D X ? - CBSE (Science) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

#### Question

If $y = \sin^{- 1} \left( 6x\sqrt{1 - 9 x^2} \right), - \frac{1}{3\sqrt{2}} < x < \frac{1}{3\sqrt{2}}$ $\frac{dy}{dx}$ ?

#### Solution

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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Solution If Y = Sin − 1 ( 6 X √ 1 − 9 X 2 ) , − 1 3 √ 2 < X < 1 3 √ 2 D Y D X ? Concept: Simple Problems on Applications of Derivatives.
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