$\displaystyle f(x) = y\cos(x) = 2x^{2} + 5y^{2}$

$\displaystyle [\cos(x)][1]y' + [y][-\sin(x)] = 4x + 10yy'$

$\displaystyle \cos(x)y' + [-\sin(x)y] = 4x + 10yy'$

$\displaystyle \cos(x)y' - \sin(x)y = 4x + 10yy'$

$\displaystyle \cos(x)y' - 10yy' = \sin(x)y + 4x$

$\displaystyle y'[\cos(x) - 10y] = \sin(x)y + 4x$

$\displaystyle y' = \dfrac{ \sin(x)y + 4x}{\cos(x) - 10y}$