Question

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

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)

Answer:

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