Question

find dy/dx at x=1,y=π/4 if sin²y+cos xy=k

Answer 1

${sin}^{2} y + \cos(xy) = k \\ differentiate \: wrt \: x\\ \\ \sin(2y) \frac{dy}{dx} - \sin(xy) (y + x \frac{dy}{dx} ) = 0 \\ \frac{dy}{dx} ( \sin(2y) - x \sin(xy) ) = y \sin(xy) \\ \frac{dy}{dx} = \frac{y \sin(xy) }{ \sin(2y) - x \sin(xy) } \\ at \: x = 1 \: and \: y = \frac{\pi}{4} \\ \frac{dy}{dx} = \frac{ \frac{\pi}{4} \sin( \frac{\pi}{4} ) }{ \sin( \frac{\pi}{2} ) - sin( \frac{\pi}{4}) } \\ \frac{dy}{dx} = \frac{ \frac{\pi}{4 \sqrt{2} } }{1 - \frac{1}{ \sqrt{2} } } \\ \frac{dy}{dx} = \frac{ \frac{\pi}{4} }{ \sqrt{2} - 1 }$
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