Find the differential of the function \cos^{-1}\left ( \sin 2x \right ) w.r.t.x.

 

 

 

 
 
 
 
 

Answers (1)

\frac{d}{dx}\left [ \cos^{-1}\left ( \sin 2x \right ) \right ]= -\frac{1}{\sqrt{1-\sin ^{2}2x}}\cdot 2\cos 2x= -\frac{2\cos 2x}{\sqrt{\cos ^{2}2x}}
                                       = \frac{-2\cos 2x}{\left | \cos 2x \right |}
\therefore \frac{d}{dx}\left [ \cos^{-1}\left ( \sin 2x \right ) \right ]= \frac{-2\cos 2x}{\cos 2x}= -2
                                              or  =   \frac{2\cos 2x}{-\cos 2x}= 2

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