Answer:
[tex]\:f'\left(x\right)=\cos \:\left(2x\right)[/tex]
Step-by-step explanation:
Given the function
[tex]\:f\left(x\right)=\left(sinx\right)cosx[/tex]
Let us take the derivative
[tex]\frac{d}{dx}\left(\sin \left(x\right)\cos \left(x\right)\right)[/tex]
Apply the product rule: [tex]\left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'[/tex]
so
[tex]\frac{d}{dx}\left(sinx\right)cosx\:=\frac{d}{dx}\left(\sin \:\left(x\right)\right)\cos \:\left(x\right)+\frac{d}{dx}\left(\cos \:\left(x\right)\right)\sin \:\left(x\right)[/tex]
as
[tex]\frac{d}{dx}\left(\sin \:\left(x\right)\right)=cos\left(x\right)[/tex]
[tex]\frac{d}{dx}\left(cos\:\left(x\right)\right)=-sin\:\left(x\right)[/tex]
so
[tex]=\cos \left(x\right)\cos \left(x\right)+\left(-\sin \left(x\right)\right)\sin \left(x\right)[/tex]
[tex]=\cos ^2\left(x\right)-\sin ^2\left(x\right)[/tex]
Use the following identity: [tex]\cos ^2\left(x\right)-\sin ^2\left(x\right)=\cos \left(2x\right)[/tex]
[tex]=\cos \left(2x\right)[/tex]
Therefore,
[tex]\:f'\left(x\right)=\cos \:\left(2x\right)[/tex]
