Question

Que7: Find dy/dx of {sin²y+cosxy=k}

Answer 1

Answer:

∴ dy/dx = ysin xy/(sin 2y-xsin xy)

Step-by-step explanation:

Given,

sin²y+cos xy = k

Differentiating both sides with respect to x,

or, d(sin²y+cos xy)/dx = dk/dx

or, dsin²y/dx + dcos xy/dx = 0 [∵ k is constant]

or, dsin²y/dsin y×dsin y/dx + dcos xy/dxy×dxy/dx = 0

or, 2sin y×cos y×dy/dx + (-sin xy)(x×dy/dx + y×dx/dx) = 0

or, sin 2y×dy/dx - xsin xy×dy/dx - ysin xy = 0

or, sin 2y×dy/dx-xsin xy×dy/dx = ysin xy

or, dy/dx(sin 2y-xsin xy) = ysin xy

∴ dy/dx = ysin xy/(sin 2y-xsin xy)