Question

find dy/dx at x=1, y= pi/4 if sin sq.y+cosxy = K

Answer 1

sin²y + cosxy = k
differentiate with respect to x
d(sin²y)/dx + d(cosxy)/dx = d(k)/dx

Let's start differentiating separately ,
d(sin²y)/dx = 2(siny)²⁻¹d(siny)/dx [∵ d(xⁿ)/dx = nxⁿ⁻¹ dx/dx ]
= 2siny.cosy .dy/dx [∵ d(sinФ)/dx = cosФ.dФ/dx ]

similarly, d(cosxy)/dx = -sinxyd(xy)/dx [ ∵d(cosФ)/dx = -sinФ.dФ/dx ]
= -sinxy.[y.dx/dx + xdy/dx ]
=- sinxy[y + x.dy/dx ]

And dK/dx = 0 [ ∵ k is constant term ]

Now, 2siny.cosy.dy/dx - sinxy[y + xdy/dx] =0
⇒dy/dx [ 2siny.cosy -xsinxy ] = ysinxy
⇒ dy/dx = ysinxy/(2siny.cosy - xsinxy )

Now, dy/dx at x = 1 and y = π/4 :-
dy/dx = π/4sinπ/4/(2sinπ/4.cosπ/4 - 1sinπ/4)
= π/4√2/(2 × 1/√2 × 1/√2 - 1/√2)
= π/4(√2 - 1)

Answer 2

This is the answer.

Last step is rationalization