How do you implicitly differentiate y /sin( 2x) = x cos(2x-y)ysin(2x)=xcos(2xy)?

1 Answer
Aug 21, 2016

y'=(sin2xcos(2x-y)+2xcos(4x-y))/(1-xsin2xsin(2x-y))

Explanation:

Let us rewrite the given eqn. as, y=xsin2xcos(2x-y).....(1).

Keeping in mind that, (uvw)'=u'vw+uv'w+uvw',

we diff. both sides of (1) w.r.t. x, to get,

y'=1*sin2xcos(2x-y)+x(2cos2x)cos(2x-y)+xsin2x(cos(2x-y))'

Here, by Chain Rule,

(cos(2x-y))'=(-sin(2x-y))(2x-y)'=(-sin(2x-y))(2-y')=(y'-2)(sin(2x-y)

Hence,

y'=sin2xcos(2x-y)+2xcos2xcos(2x-y)+x(y'-2)sin2xsin(2x-y)

=sin2xcos(2x-y)+2xcos2xcos(2x-y)

+xy'sin2xsin(2x-y)-2xsin2xsin(2x-y)

:. y'(1-xsin2xsin(2x-y))=sin2xcos(2x-y)+2x{cos2xcos(2x-y)-sin2xsin(2x-y)}

=sin2xcos(2x-y)+2x{cos(2x+(2x-y))}

=sin2xcos(2x-y)+2xcos(4x-y)

Therefore,

y'=(sin2xcos(2x-y)+2xcos(4x-y))/(1-xsin2xsin(2x-y))

Enjoy Maths.!