Use the Product Rule to find the derivative \( \frac{d}{d x}\left[\left(\frac{1}{3} x^{3}+\frac{2}{5} x^{5}\right)\left(\frac{2}{x}-4 x^{2}\right)\right] \)

To find the derivative of the expression \[ y = \left(\frac{1}{3} x^{3}+\frac{2}{5} x^{5}\right)\left(\frac{2}{x}-4 x^{2}\right) \] using the Product Rule, we will follow these steps: 1. **Identify the two functions**: - Let \( u = \frac{1}{3} x^{3} + \frac{2}{5} x^{5} \) - Let \( v = \frac{2}{x} - 4 x^{2} \) 2. **Differentiate each function**: - We need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \). 3. **Apply the Product Rule**: - The Product Rule states that \( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \). Now, let's calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) step by step. ### Step 1: Differentiate \( u \) \[ u = \frac{1}{3} x^{3} + \frac{2}{5} x^{5} \] Using the power rule: \[ \frac{du}{dx} = \frac{1}{3} \cdot 3 x^{2} + \frac{2}{5} \cdot 5 x^{4} = x^{2} + 2 x^{4} \] ### Step 2: Differentiate \( v \) \[ v = \frac{2}{x} - 4 x^{2} \] This can be rewritten as \( v = 2x^{-1} - 4x^{2} \). Now, differentiating: \[ \frac{dv}{dx} = -2x^{-2} - 8x = -\frac{2}{x^{2}} - 8x \] ### Step 3: Apply the Product Rule Now we can apply the Product Rule: \[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \] Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \): \[ \frac{dy}{dx} = \left(\frac{1}{3} x^{3} + \frac{2}{5} x^{5}\right)\left(-\frac{2}{x^{2}} - 8x\right) + \left(\frac{2}{x} - 4x^{2}\right)\left(x^{2} + 2x^{4}\right) \] Now, we can simplify this expression to find the final derivative. Let's calculate this expression step by step. 1. Calculate \( u \frac{dv}{dx} \) 2. Calculate \( v \frac{du}{dx} \) 3. Combine the results. Let's perform these calculations. ### Step 1: Calculate \( u \frac{dv}{dx} \) \[ u \frac{dv}{dx} = \left(\frac{1}{3} x^{3} + \frac{2}{5} x^{5}\right)\left(-\frac{2}{x^{2}} - 8x\right) \] ### Step 2: Calculate \( v \frac{du}{dx} \) \[ v \frac{du}{dx} = \left(\frac{2}{x} - 4x^{2}\right)\left(x^{2} + 2x^{4}\right) \] Now, let's compute these two products. ### Step 1 Calculation \[ u \frac{dv}{dx} = \left(\frac{1}{3} x^{3} + \frac{2}{5} x^{5}\right)\left(-\frac{2}{x^{2}} - 8x\right) \] Expanding this: \[ = -\frac{2}{x^{2}} \cdot \frac{1}{3} x^{3} - 8x \cdot \frac{1}{3} x^{3} - \frac{2}{x^{2}} \cdot \frac{2}{5} x^{5} - 8x \cdot \frac{2}{5} x^{5} \] \[ = -\frac{2}{3} x + -\frac{8}{3} x^{4} - \frac{4}{5} x^{3} - \frac{16}{5} x^{6} \] ### Step 2 Calculation \[ v \frac{du}{dx} = \left(\frac{2}{x} - 4x^{2}\right)\left(x^{2} + 2x^{4}\right) \] Expanding this: \[ = \frac{2}{x} \cdot x^{2} + \frac{2}{x} \cdot 2x^{4} - 4x^{2} \cdot x^{2} - 4x^{2} \cdot 2x^{4} \] \[ = 2 + 4x^{3} - 4x^{4} - 8x^{6} \] ### Combine Results Now, we combine both results: \[ \frac{dy}{dx} = \left(-\frac{2}{3} x - \frac{8}{3} x^{4} - \frac{4}{5} x^{3} - \frac{16}{5} x^{6}\right) + \left(2 + 4x^{3} - 4x^{4} - 8x^{6}\right) \] Combining like terms will give us the final derivative. Let's simplify this expression. 1. Combine constant terms. 2. Combine \( x \) terms. 3. Combine \( x^{2} \) terms. 4. Combine \( x^{3} \) terms. 5. Combine \( x^{4} \) terms. 6. Combine \( x^{5} \) terms. 7. Combine \( x^{6} \) terms. After combining, we will have the final derivative. Let's perform the final simplification. ### Final Derivative Combining all terms, we get: \[ \frac{dy}{dx} = \left(2 - \frac{2}{3} x\right) + \left(4 - \frac{4}{5}\right)x^{3} + \left(-\frac{8}{3} - 4\right)x^{4} + \left(-\frac{16}{5} - 8\right)x^{6} \] This will yield the final expression for the derivative. If you need the