How is the derivative of $x\cos x\sin x$ $\frac12\sin2x + x\cos2x$?

Question

Find $f'(x)$ when $f(x) = x\cos(x) \sin(x)$

I tried using the product rule on $x\cos x$ and got: $$(x\cos x)' = -x\sin x + \cos x$$ then I used the product rule to differentiate $x\cos x\sin x$ and got: $$(x\cos x \sin x)' = (-x \sin x + \cos x)\sin x+ (x\cos x \cos x)= -x\sin^2x+ \sin x \cos x + x\cos^2x$$ My testbook says the answer is $\frac12\sin2x+ x\cos2x$

Can someone please explain how to achieve the answer shown in my textbook? I am new to calc and trig, and I'm very confused. Any help is appreciated.

Answer 1

You're right. Use,

$2\sin x\cos x = \sin(2x)$

and

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

So that,

$x(\cos^2x-\sin^2x) + \sin x \cos x = x\cos(2x) + \frac{1}{2}\sin(2x)$

Answer 2

The two answers are equivalent: $$-x\sin^2x+\sin x\cos x+x\cos^2x=x(\cos^2x-\sin^2x)+\frac12\sin2x=x\cos2x+\frac12\sin2x$$