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

Question

Elijah Abram · Accepted Answer

#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))'#

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.!

Aurora Alvarado · Answer

undefined To implicitly differentiate the equation $ \frac{y}{\sin(2x)} = x \cos(2x-y) $, follow these steps:

1. Differentiate each term of the equation with respect to $ x $.
2. Use the quotient rule to differentiate $ \frac{y}{\sin(2x)} $ and the chain rule to differentiate $ \cos(2x-y) $.
3. Isolate $ \frac{dy}{dx} $ by solving for it.

After applying these steps, the result of the implicit differentiation will yield $ \frac{dy}{dx} $ in terms of $ x $ and $ y $.