1. We need to find the derivative of the function $$y=\frac{1}{x}\sin^2(x) - 5x - \frac{x}{3}\cos^3(x).$$ 2. First, rewrite the function clearly as $$y=\frac{\sin^2(x)}{x} - 5x - \frac{x}{3}\cos^3(x).$$ 3. We will use the quotient rule and product rule, along with chain rule. 4. Derivative of the first term $$\frac{\sin^2(x)}{x}$$: - Let $$u=\sin^2(x)$$ and $$v=x$$. - Then $$u' = 2\sin(x)\cos(x)$$ by chain rule, and $$v' = 1$$. - By quotient rule, $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} = \frac{2\sin(x)\cos(x) \cdot x - \sin^2(x) \cdot 1}{x^2} = \frac{2x\sin(x)\cos(x) - \sin^2(x)}{x^2}.$$ 5. Derivative of second term $$-5x$$ is simply $$-5$$. 6. Derivative of third term $$-\frac{x}{3} \cos^3(x)$$: - Write as $$-\frac{1}{3} x \cos^3(x)$$. - Use product rule: derivative is $$-\frac{1}{3} \left(1 \cdot \cos^3(x) + x \cdot 3 \cos^2(x)(-\sin(x)) \right)$$ by chain rule. - Simplify: $$-\frac{1}{3} (\cos^3(x) - 3x \cos^2(x)\sin(x)) = -\frac{1}{3} \cos^3(x) + x \cos^2(x) \sin(x).$$ 7. Combine all derivatives: $$\frac{dy}{dx} = \frac{2x\sin(x)\cos(x) - \sin^2(x)}{x^2} - 5 - \frac{1}{3} \cos^3(x) + x \cos^2(x) \sin(x).$$ This is the derivative of the given function.