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Question
If \[\sin^2 y + \cos xy = k,\] find \[\frac{dy}{dx}\] at \[x = 1 , \] \[y = \frac{\pi}{4} .\]
Solution
We have , \[\sin^2 y + \cos xy = k\]
From the given eq.
`2 siny cos y. dy/dx - sin xy.[x.dy/dx + y.1] = 0`
or
`dy/dx = (ysinxy)/(sin2y - xsin(xy))`
`dy/dx = -(ysinxy)/(sin2y - xsinxy)`
`(dy/dx)_(y = pi/4)^(x = 1) = (pi/4 sinx1.pi/4)/(sin2(xpi)/4 - 1sin1.pi/4)`
= `(pi/4 . 1/sqrt(2))/(1 - 1/sqrt(2)) = pi/(4(sqrt(2) - 1))`
∴ `dy/dx|_(x = 1 , y = pi/4) = pi/(4(sqrt(2) - 1))`
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