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Question
Find \[\frac{dy}{dx}\] in the following case \[\sin xy + \cos \left( x + y \right) = 1\] ?
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Solution
\[\text{ We have, } \sin x y + \cos\left( x + y \right) = 1\]
Differentiating with respect to x, we get,
\[\frac{d}{dx}\left( \sin xy \right) + \frac{d}{dx}\cos\left( x + y \right) = \frac{d}{dx}\left( 1 \right)\]
\[ \Rightarrow \cos xy\frac{d}{dx}\left( xy \right) - \sin\left( x + y \right)\frac{d}{dx}\left( x + y \right) = 0 \]
\[ \Rightarrow \cos xy\left[ x\frac{d y}{d x} + y\frac{d}{dx}\left( x \right) \right] - \sin\left( x + y \right)\left[ 1 + \frac{d y}{d x} \right] = 0\]
\[ \Rightarrow \cos xy\left[ x\frac{d y}{d x} + y\left( 1 \right) \right] - \sin\left( x + y \right) - \sin\left( x + y \right)\frac{d y}{d x} = 0\]
\[ \Rightarrow x\cos xy\frac{d y}{d x} + y\cos xy - \sin\left( x + y \right) - \sin\left( x + y \right)\frac{d y}{d x} = 0\]
\[ \Rightarrow \left[ x\cos xy - \sin\left( x + y \right) \right]\frac{d y}{d x} = \left[ \sin\left( x + y \right) - y\cos xy \right]\]
\[ \Rightarrow \frac{d y}{d x} = \left[ \frac{\sin\left( x + y \right) - y \cos xy}{x\cos xy - \sin\left( x + y \right)} \right]\]
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