Question

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Answer 1

Answer:

$\:f'\left(x\right)=\cos \:\left(2x\right)$                

Step-by-step explanation:

Given the function

$\:f\left(x\right)=\left(sinx\right)cosx$

Let us take the derivative

$\frac{d}{dx}\left(\sin \left(x\right)\cos \left(x\right)\right)$

Apply the product rule:   $\left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'$

so

$\frac{d}{dx}\left(sinx\right)cosx\:=\frac{d}{dx}\left(\sin \:\left(x\right)\right)\cos \:\left(x\right)+\frac{d}{dx}\left(\cos \:\left(x\right)\right)\sin \:\left(x\right)$

as

$\frac{d}{dx}\left(\sin \:\left(x\right)\right)=cos\left(x\right)$

$\frac{d}{dx}\left(cos\:\left(x\right)\right)=-sin\:\left(x\right)$

so

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

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

Use the following identity:  $\cos ^2\left(x\right)-\sin ^2\left(x\right)=\cos \left(2x\right)$

                      $=\cos \left(2x\right)$

Therefore,

$\:f'\left(x\right)=\cos \:\left(2x\right)$