To solve for [tex]\( f^{\prime \prime \prime}(\pi) \)[/tex] given the function [tex]\( f(x) = 2 \sin x \cos x + \frac{1}{3} x^3 - x^2 \)[/tex], let's go through the differentiation process step-by-step.
Given:
[tex]\[ f(x) = 2 \sin x \cos x + \frac{1}{3} x^3 - x^2 \][/tex]
1. First Step: Simplify the Trigonometric Part
Using the double angle identity, [tex]\(2 \sin x \cos x = \sin(2x)\)[/tex], the function can be rewritten:
[tex]\[ f(x) = \sin(2x) + \frac{1}{3} x^3 - x^2 \][/tex]
2. First Derivative [tex]\( f'(x) \)[/tex]
Compute the first derivative:
[tex]\[ f'(x) = \frac{d}{dx}(\sin(2x)) + \frac{d}{dx}\left(\frac{1}{3} x^3\right) - \frac{d}{dx}(x^2) \][/tex]
Using the chain rule and basic differentiation rules:
[tex]\[ \frac{d}{dx}(\sin(2x)) = 2 \cos(2x) \][/tex]
[tex]\[ \frac{d}{dx}\left(\frac{1}{3} x^3\right) = x^2 \][/tex]
[tex]\[ \frac{d}{dx}(x^2) = 2x \][/tex]
So:
[tex]\[ f'(x) = 2 \cos(2x) + x^2 - 2x \][/tex]
3. Second Derivative [tex]\( f''(x) \)[/tex]
Compute the second derivative:
[tex]\[ f''(x) = \frac{d}{dx}(2 \cos(2x) + x^2 - 2x) \][/tex]
Using the chain rule and basic differentiation rules:
[tex]\[ \frac{d}{dx}(2 \cos(2x)) = -4 \sin(2x) \][/tex]
[tex]\[ \frac{d}{dx}(x^2) = 2x \][/tex]
[tex]\[ \frac{d}{dx}(-2x) = -2 \][/tex]
So:
[tex]\[ f''(x) = -4 \sin(2x) + 2x - 2 \][/tex]
4. Third Derivative [tex]\( f'''(x) \)[/tex]
Compute the third derivative:
[tex]\[ f'''(x) = \frac{d}{dx}(-4 \sin(2x) + 2x - 2) \][/tex]
Using the chain rule and basic differentiation rules:
[tex]\[ \frac{d}{dx}(-4 \sin(2x)) = -8 \cos(2x) \][/tex]
[tex]\[ \frac{d}{dx}(2x) = 2 \][/tex]
[tex]\[ \frac{d}{dx}(-2) = 0 \][/tex]
So:
[tex]\[ f'''(x) = -8 \cos(2x) + 2 \][/tex]
5. Evaluate at [tex]\( x = \pi \)[/tex]
Substitute [tex]\( x = \pi \)[/tex] into the third derivative:
[tex]\[ f'''(\pi) = -8 \cos(2\pi) + 2 \][/tex]
Recall that [tex]\( \cos(2\pi) = 1 \)[/tex]:
[tex]\[ f'''(\pi) = -8 \times 1 + 2 = -8 + 2 = -6 \][/tex]
So:
[tex]\[ f'''(\pi) = -6 \][/tex]
Hence, the correct value is:
[tex]\[ \boxed{-6} \][/tex]
