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Explain solution RD Sharma class 12 chapter Differentiation exercise 10.5 question 48 maths

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Answer: \frac{d y}{d x}=\frac{1-(x+y) y \cot u}{(x+y) \log \sin x-1}

Hint: To solve this we convert terms into log

Given: (\sin x)^{y}=x+y

Solution:  

        (\sin x)^{y}=x+y                                    \left[\because \frac{d y}{d x} \int \log \sin x=y\right]

        \log (\sin x)^{y}=\log (x+y)                    \left[y=\frac{1}{\sin x} \times \cos x\right]

        y \log (\sin x)=\log (x+y)                    \left[\frac{\cos x}{\sin x}=\cot x\right]

        y \frac{d}{d x} \log (\sin x)+\log \sin x \cdot \frac{d y}{d x}=\frac{1}{x+y} \frac{d}{d x}(x+y)

        y \cot x+\log \sin x \cdot \frac{d y}{d x}=\frac{1}{x+y}+\frac{1}{x+y} \frac{d y}{d x}

        \frac{d y}{d x}\left(\frac{1}{x+y}-\log \sin x\right)=y \cot x \cdot \frac{-1}{x+y}

        \frac{d y}{d x}\left(\frac{1-(x+y) \log \sin x}{x+y}\right)=\frac{(x+y) y \cot x-1}{x+y}

        \begin{aligned} &\frac{d y}{d x}=\frac{(x+y) y \cot x-1}{1-(x+y) \log \sin x} \\\\ &\frac{d y}{d x}=\frac{-(1-y \cot x(x+y)}{-(x+y) \log \sin x-1} \end{aligned}

        d y=\frac{1-(x+y) \cdot y \cot x}{(x+y) \log \sin x-1}

Hence proved

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