Question

Find $\frac {dy}{dx}:$ $sin^2x+cos^2y=1$

Answer 1

The given relationship is sin2x + cos2y = 1
Differentiating this relationship with respect to x, we obtain
$\frac {d}{dx}$(sin2x + cos2y) = $\frac {d}{dx}$(1)

\implies$$\frac {d}{dx}(sin2x) + $\frac {d}{dx}$(cos2y) = 0

$\implies$2sin x . $\frac {d}{dx}$(sin x) + 2cos y . $\frac {d}{dx}$(cos y) = 0

$\implies$2sin x . cos x + 2cos y (-sin y) . $\frac {dy}{dx}$ = 0

$\implies$sin2x - sin2y$\frac {dy}{dx}$ = 0

∴$\frac {dy}{dx}$ = $\frac {sin\ 2x} {sin\ 2y}$