Answer:
2cosx - 2xsinx
Step-by-step explanation:
Differentiate using the product rule
Given y = f(x)g(x), then
[tex]\frac{dy}{dx}[/tex] = f(x)g'(x) + g(x)f'(x)
Here
f(x) = 2x ⇒ f'(x) = 2
g(x) = cos x ⇒ g'(x) = - sin x
Hence
y = 2x cos x
[tex]\frac{dy}{dx}[/tex] = 2x(- sinx) + 2cosx = 2cos x - 2xsin x