Following are the calculation to the method value:
Given:
[tex]\cos (xy) = y-1\\\\ x=\frac{\pi}{2} \ \ and \ \ y = 1[/tex]
To find:
[tex]\frac{dy}{dx}=?[/tex]
Solution:
[tex]\to \cos (xy) = y-1\\\\\to \frac{d}{x} \ (\cos xy) = \frac{d}{dx}\ (y-1)\\\\\to -\ ( \sin xy) \frac{d}{dx} \ (xy) = \frac{dy}{dx}\\\\\to -\ ( \sin xy) ( x\ \frac{d}{dx} + y) = \frac{dy}{dx}\\\\\to - x \sin (xy) \ \frac{dy}{dx} -y\sin (xy) =\frac{dy}{dx}\\\\\to \frac{dy}{dx} (1+ x \sin \ xy) = - y \sin xy\\\\\to \frac{dy}{dx}=\frac{-y \sin xy}{1+x\sin xy }\\\\[/tex]
When [tex]\ x=\frac{\pi }{2} \ \ \ and \ \ \ y=1\\\\[/tex]
[tex]\to \frac{dy}{dx}=\frac{- 1 \sin (\frac{\pi }{2})}{1+ \frac{\pi }{2} \sin \frac{\pi }{2} }\\\\\to \frac{dy}{dx}=\frac{- 1 }{1+ \frac{\pi }{2} } = \frac{-1 }{\frac{2+\pi }{2}}\\\\\to \frac{dy}{dx}=\frac{-2}{2+\pi }\\\\[/tex]
Learn more:
brainly.com/question/23143484
