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Solution for Fine D Y D X Y = ( Sin X ) X + Sin − 1 √ X ? - CBSE (Science) Class 12 - Mathematics

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Question

Fine \[\frac{dy}{dx}\] \[y = \left( \sin x \right)^x + \sin^{- 1} \sqrt{x}\] ?

Solution

\[\text{ We have, y }= \left( \sin x \right)^x + \sin^{- 1} \sqrt{x}\]

\[ \Rightarrow y = e^{\log \left( \sin x \right)^x} + \sin^{- 1} \sqrt{x}\]

\[ \Rightarrow y = e^{x\log \sin x} + \sin^{- 1} \sqrt{x}\]

Differentiating with respect to using chain rule,

\[\frac{dy}{dx} = \frac{d}{dx}\left( e^{x \log \sin x} \right) + \frac{d}{dx}\left( \sin^{- 1} \sqrt{x} \right)\]

\[ = e^{x \log \sin x} \frac{d}{dx}\left( x \log \sin x \right) + \frac{1}{\sqrt{1 - \left( \sqrt{x} \right)^2}}\frac{d}{dx}\left( \sqrt{x} \right)\]

\[ = e^{ \log \left( \sin x \right)^x} \left[ x\frac{d}{dx}\left( \log \sin x \right) + \log \sin x\frac{d}{dx}\left( x \right) + \frac{1}{\sqrt{1 - x}} \times \frac{1}{2\sqrt{x}} \right] \]

\[ = \left( \sin x \right)^x \left[ x\frac{1}{\sin x}\frac{d}{dx}\left( \sin x \right) + \log \sin x \right] + \frac{1}{2\sqrt{x - x^2}}\]

\[ = \left( \sin x \right)^x \left[ \frac{x}{\sin x}\left( \cos x \right) + \log \sin x \right] + \frac{1}{2\sqrt{x - x^2}}\]

\[ = \left( \sin x \right)^x \left[ x\cot x + \log \sin x \right] + \frac{1}{2\sqrt{x - x^2}}\]

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Solution for question: Fine D Y D X Y = ( Sin X ) X + Sin − 1 √ X ? concept: Simple Problems on Applications of Derivatives. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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