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Question
If \[\sin \left( x + y \right) = \log \left( x + y \right), \text { then } \frac{dy}{dx} =\] ___________ .
Options
2
-2
1
-1
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Solution
− 1
\[\text { We have }, \sin\left( x + y \right) = \log\left( x + y \right)\]
\[ \Rightarrow \cos\left( x + y \right)\left( 1 + \frac{dy}{dx} \right) = \frac{1}{\left( x + y \right)}\left( 1 + \frac{dy}{dx} \right)\]
\[ \Rightarrow \cos\left( x + y \right) + \cos\left( x + y \right)\frac{dy}{dx} = \frac{1}{\left( x + y \right)} + \frac{1}{\left( x + y \right)}\frac{dy}{dx}\]
\[ \Rightarrow \cos\left( x + y \right)\frac{dy}{dx} - \frac{1}{\left( x + y \right)}\frac{dy}{dx} = \frac{1}{\left( x + y \right)} - \cos\left( x + y \right)\]
\[ \Rightarrow \left\{ \cos\left( x + y \right) - \frac{1}{\left( x + y \right)} \right\}\frac{dy}{dx} = \frac{1}{\left( x + y \right)} - \cos\left( x + y \right)\]
\[ \Rightarrow - \left\{ \frac{1}{\left( x + y \right)} - \cos\left( x + y \right) \right\}\frac{dy}{dx} = \frac{1}{\left( x + y \right)} - \cos\left( x + y \right)\]
\[ \Rightarrow \frac{dy}{dx} = - 1\]
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