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If Sin ( X + Y ) = Log ( X + Y ) , Then D Y D X = (A) 2 (B) − 2 (C) 1 (D) − 1] - Mathematics

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प्रश्न

If \[\sin \left( x + y \right) = \log \left( x + y \right), \text { then } \frac{dy}{dx} =\] ___________ .

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MCQ

उत्तर

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\[\text { We have }, \sin\left( x + y \right) = \log\left( x + y \right)\]
\[ \Rightarrow \cos\left( x + y \right)\left( 1 + \frac{dy}{dx} \right) = \frac{1}{\left( x + y \right)}\left( 1 + \frac{dy}{dx} \right)\]
\[ \Rightarrow \cos\left( x + y \right) + \cos\left( x + y \right)\frac{dy}{dx} = \frac{1}{\left( x + y \right)} + \frac{1}{\left( x + y \right)}\frac{dy}{dx}\]
\[ \Rightarrow \cos\left( x + y \right)\frac{dy}{dx} - \frac{1}{\left( x + y \right)}\frac{dy}{dx} = \frac{1}{\left( x + y \right)} - \cos\left( x + y \right)\]
\[ \Rightarrow \left\{ \cos\left( x + y \right) - \frac{1}{\left( x + y \right)} \right\}\frac{dy}{dx} = \frac{1}{\left( x + y \right)} - \cos\left( x + y \right)\]
\[ \Rightarrow - \left\{ \frac{1}{\left( x + y \right)} - \cos\left( x + y \right) \right\}\frac{dy}{dx} = \frac{1}{\left( x + y \right)} - \cos\left( x + y \right)\]
\[ \Rightarrow \frac{dy}{dx} = - 1\]

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अध्याय 11: Differentiation - Exercise 11.10 [पृष्ठ १२०]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 11 Differentiation
Exercise 11.10 | Q 13 | पृष्ठ १२०

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