Question

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

Options

• 2

• -2

• 1

• -1

Answer 1

− 1

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