The correct answer is \(=(a\,sin\,x-b\,cos\,x).sin(a\,cosx+b\,sinx)\)
Let \(y=cos(a\,cosx+b\,sinx)\)
By using chain rule, we obtain
\(\frac{dy}{dx}=\frac{d}{dx}cos(a\,cosx+b\,sinx)\)
\(⇒\frac{dy}{dx}=-sin(a\,cosx+b\,sinx).\frac{d}{dx}(a\,cosx+b\,sinx)\)
\(=-sin(a\,cosx+b\,sinx).[a(-sinx)+b\,cosx]\)
\(=(a\,sin\,x-b\,cos\,x).sin(a\,cosx+b\,sinx)\)