Question

If \[\sin^2 y + \cos xy = k,\] find  \[\frac{dy}{dx}\] at \[x = 1 , \] \[y = \frac{\pi}{4} .\] 

Answer 1

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))`