Question

Find \(\frac {dy}{dx}\):  \(ax+by^2=cos\ y\)

Answer 1

The given relationship is ax+by² = cos y
Differentiating this relationship with respect to x, we obtain
\(\frac {d}{dx}\)(ax+by²) = \(\frac {d}{dx}\)(cos y)

\(\implies\)\(\frac {d}{dx}\)(ax)+\(\frac {d}{dx}\)(by²) = \(\frac {d}{dx}\)(cosy)

\(\implies\)a+b\(\frac {d}{dx}\)(y²) = \(\frac {d}{dx}\)(cos y)         …...…….(1)

Using chain rule,we obtain \(\frac {d}{dx}\)(y²) = 2y\(\frac {dy}{dx}\) and \(\frac {d}{dx}\)(cosy) = -siny \(\frac {dy}{dx}\)           ………....(2)
From (1) and (2), we obtain
a+b.2y\(\frac {dy}{dx}\) = -sin y.\(\frac {dy}{dx}\)

\(\implies\)(2by+sin y)\(\frac {dy}{dx}\) = -a

∴ \(\frac {dy}{dx}\) = \(\frac {-a}{2by+sin\ y}\)