Question

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Answer 1

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

Explanation:

Given the function

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

Let us take the derivative

`(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

`(d)/(dx)\left(sinx\right)cosx\:=(d)/(dx)\left(\sin \:\left(x\right)\right)\cos \:\left(x\right)+(d)/(dx)\left(\cos \:\left(x\right)\right)\sin \:\left(x\right)`

as

`(d)/(dx)\left(\sin \:\left(x\right)\right)=cos\left(x\right)`

`(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)`