Answer:
[tex]\sf -5\cos \left(x\right)+6\cos ^3\left(x\right)[/tex]
explanation:
[tex]\sf y = sin(x) * cos(2x)[/tex]
[tex]\rightarrow \sf \frac{d}{dx}\left(sin\left(x\right)\ * \:\:cos\left(2x\right)\right)[/tex]
[tex]\sf \bold {Apply\:the\:Product\:Rule}:\quad \left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'[/tex]
[tex]\rightarrow \sf \frac{d}{dx}\left(\sin \left(x\right)\right)\cos \left(2x\right)+\frac{d}{dx}\left(\cos \left(2x\right)\right)\sin \left(x\right)[/tex]
[tex]\sf \bold{ Apply \ differentiation \ rule \ \ \ : } \ \ \ sin(x) = cos(x) \ \ and \ \ cos(x) = -sin(x)[/tex]
[tex]\rightarrow \sf \cos \left(x\right)\cos \left(2x\right)+\left(-\sin \left(2x\right)\ * \:2\right)\sin \left(x\right)[/tex]
[tex]\rightarrow \sf \cos \left(x\right)\cos \left(2x\right)\left-2\sin \left(2x\right)\sin \left(x\right)[/tex]
[tex]\sf \bold {use \ the \ formulae \ : \ cos(2x) = 2cos^2(x) - 1} \ {and} \ \ \sf \bold{sin(x) = 2 sin x cos x}[/tex]
[tex]\rightarrow \sf cos(x) (2cos^2 (x) -1) -2(2sin(x)cos(x)sin(x))[/tex]
[tex]\rightarrow \sf 2cos^3 (x) - cos(x) - 4sin^2(x) cos(x)[/tex]
[tex]\rightarrow \sf -5\cos \left(x\right)+6\cos ^3\left(x\right)[/tex]
