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

Aug 21, 2016

$y ' = \frac{\sin 2 x \cos \left(2 x - y\right) + 2 x \cos \left(4 x - y\right)}{1 - x \sin 2 x \sin \left(2 x - y\right)}$

#### Explanation:

Let us rewrite the given eqn. as, $y = x \sin 2 x \cos \left(2 x - y\right) \ldots . . \left(1\right)$.

Keeping in mind that, $\left(u v w\right) ' = u ' v w + u v ' w + u v w '$,

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

$y ' = 1 \cdot \sin 2 x \cos \left(2 x - y\right) + x \left(2 \cos 2 x\right) \cos \left(2 x - y\right) + x \sin 2 x \left(\cos \left(2 x - y\right)\right) '$

Here, by Chain Rule,

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

Hence,

$y ' = \sin 2 x \cos \left(2 x - y\right) + 2 x \cos 2 x \cos \left(2 x - y\right) + x \left(y ' - 2\right) \sin 2 x \sin \left(2 x - y\right)$

$= \sin 2 x \cos \left(2 x - y\right) + 2 x \cos 2 x \cos \left(2 x - y\right)$

$+ x y ' \sin 2 x \sin \left(2 x - y\right) - 2 x \sin 2 x \sin \left(2 x - y\right)$

$\therefore y ' \left(1 - x \sin 2 x \sin \left(2 x - y\right)\right) = \sin 2 x \cos \left(2 x - y\right) + 2 x \left\{\cos 2 x \cos \left(2 x - y\right) - \sin 2 x \sin \left(2 x - y\right)\right\}$

$= \sin 2 x \cos \left(2 x - y\right) + 2 x \left\{\cos \left(2 x + \left(2 x - y\right)\right)\right\}$

$= \sin 2 x \cos \left(2 x - y\right) + 2 x \cos \left(4 x - y\right)$

Therefore,

$y ' = \frac{\sin 2 x \cos \left(2 x - y\right) + 2 x \cos \left(4 x - y\right)}{1 - x \sin 2 x \sin \left(2 x - y\right)}$

Enjoy Maths.!