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

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Answer 1 · 2016-08-21T11:13:57

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

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,

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

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

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.!

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