Question

Find dy/dx, if cos(xy) = x+ y​

Answer 1

Question :-

$\sf{Find \: \: \frac{dy}{dx} \: \: \: if \: \cos(xy) = x + y}$

Answer :-

$\rightarrow \: \boxed{\sf{ \frac{dy}{dx} = - \frac{1 +y \sin(xy) }{1 + x \sin(xy)} }} \:$

Solution :-

According to the question,

$\rightarrow \sf{ \cos \: (xy) = x + y} \\ \\ \text{differentiating \: with \: respect \: to \: x \: } \\ \\ \rightarrow \sf{ - \sin(xy) \: \bigg(x \frac{dy}{dx} + y \bigg) = 1 + \frac{dy}{dx} } \\ \\ \rightarrow \sf{- x \: \sin(xy) \: \frac{dy}{dx} - y \sin(xy) = 1 + \frac{dy}{dx} } \\ \\ \rightarrow \sf{ \frac{dy}{dx} + x \sin(xy) \: \frac{dy}{dx} = - 1 - y \sin \: xy)} \\ \\ \rightarrow \boxed{\sf{ \frac{dy}{dx} = - \frac{1 +y \sin(xy) }{1 + x \sin(xy)} }}$

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Answer 2

$\bold{\large{\underline{\underline{\sf{StEp\:by\:stEp\:explanation:}}}}}$

According to the question, dy/dx, if cos(xy) = x+ y

$\begin{lgathered}\rightarrow \tt{ \cos \: (xy) = x + y} \\ \\ \red{differentiating \: with \: respect \: to \: x \: } \\ \\ \rightarrow \tt{ - \sin(xy) \: \bigg(x \frac{dy}{dx} + y \bigg) = 1 + \frac{dy}{dx} } \\ \\ \rightarrow \tt{- x \: \sin(xy) \: \frac{dy}{dx} - y \sin(xy) = 1 + \frac{dy}{dx} } \\ \\ \rightarrow \tt{ \frac{dy}{dx} + x \sin(xy) \: \frac{dy}{dx} = - 1 - y \sin \: xy)} \\ \\ \rightarrow \boxed{\tt{ \frac{dy}{dx} = - \frac{1 +y \sin(xy) }{1 + x \sin(xy)} }}\end{lgathered}$