Question

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

Answer 1

The given relationship is sin2y + cos xy = $\pi$

Differentiating this relationship with respect to x, we obtain

$\frac {d}{dx}$(sin2y + cos xy) = $\frac {d}{dx}$($\pi$)

⇒$\frac {d}{dx}$(sin2y) + $\frac {d}{dx}$ (cos xy) = 0 ……..... (1)
Using chain rule, we obtain

\implies$$\frac {d}{dx}(sin2y) = 2sin y . $\frac {d}{dx}$(sin y) = 2sin y cos y $\frac {dy}{dx}$ ……...... (2)

$\frac {d}{dx}$(cos xy) = -sin xy. $\frac {d}{dx}$(xy) = - sin xy [y$\frac {d}{dx}$(x) + x$\frac {d}{dx}$]

= -sin xy [y.1+x$\frac {dy}{dx}$] = -y sin xy - x sin xy . $\frac {dy}{dx}$ ………..... (3)

From (1), (2) and (3), we obtain

2sin y cos y $\frac {dy}{dx}$ - y sin xy - x sin xy $\frac {dy}{dx}$ = 0

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

$\implies$(sin 2y - x sin xy)$\frac {dy}{dx}$ = y sin xy

∴$\frac {dy}{dx}$ = $\frac {y sin \ xy}{sin\ 2y-xsin\ xy}$