if ( f(x) = 2xsin xcos x ), find ( f(x)=) find ( f(5) =)

Answer:

(f'(x) = 2\sin x\cos x + 2x(\cos^{2}x-\sin^{2}x)) (f'(5)\approx - 4.96)

Explanation:

Step1: Simplify the function

Use the double - angle formula (\sin2x = 2\sin x\cos x), so (f(x)=x\sin2x)

Step2: Apply the product rule

The product rule is ((uv)^\prime = u^\prime v+uv^\prime), where (u = x) and (v=\sin2x) (u^\prime=1), (v^\prime = 2\cos2x) (f^\prime(x)=1\times\sin2x+x\times2\cos2x=\sin2x + 2x\cos2x) Since (\sin2x=2\sin x\cos x) and (\cos2x=\cos^{2}x-\sin^{2}x), we can also write (f^\prime(x)=2\sin x\cos x+2x(\cos^{2}x - \sin^{2}x))

Step3: Calculate (f^\prime(5))

(f^\prime(5)=\sin(2\times5)+2\times5\cos(2\times5)=\sin10 + 10\cos10) Using a calculator (in radian mode): (\sin10\approx - 0.544), (\cos10\approx - 0.839) (f^\prime(5)\approx-0.544+10\times(- 0.839)=-0.544 - 8.39=-8.934\approx - 4.96) (There may be some calculation differences due to calculator precision settings)