Let's solve the differentiation step by step. Given: \[ y = \sqrt[3]{x^3 + 15x + 4} \cdot x^4 \] --- ### **Step 1: Apply the Product Rule** Let \( u = \sqrt[3]{x^3 + 15x + 4} \) and \( v = x^4 \). The product rule states: \[ y' = u'v + uv' \] **Explanation:** We broke the function into two parts to use the product rule for differentiation. --- ### **Step 2: Find \( u' \) and \( v' \), Then Substitute** **Differentiate \( v \):** \[ v = x^4 \implies v' = 4x^3 \] **Differentiate \( u \):** \[ u = (x^3 + 15x + 4)^{1/3} \] Using the chain rule: \[ u' = \frac{1}{3}(x^3 + 15x + 4)^{-2/3} \cdot (3x^2 + 15) \] Now plug into the product rule: \[ y' = u'v + uv' \] \[ = \left[\frac{1}{3}(x^3 + 15x + 4)^{-2/3}(3x^2 + 15)\right] x^4 + (x^3 + 15x + 4)^{1/3} \cdot 4x^3 \] **Explanation:** We used the chain rule to differentiate \( u \) and the power rule for \( v \), then plugged them into the product rule formula. --- ## **Final Answer:** \[ \boxed{ y' = \frac{1}{3}(x^3 + 15x + 4)^{-2/3}(3x^2 + 15)x^4 + 4x^3(x^3 + 15x + 4)^{1/3} } \]