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

ConceptSimple Problems on Applications of Derivatives

#### 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 Fine D Y D X Y = ( Sin X ) X + Sin − 1 √ X ? Concept: Simple Problems on Applications of Derivatives.
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