1. **Problem:** Differentiate the function $$h(x) = (2x + 5)^7 (3x^4 - 8)^5$$. 2. **Formula and rules:** To differentiate a product of two functions, use the product rule: $$\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$$ where $$u(x) = (2x + 5)^7$$ and $$v(x) = (3x^4 - 8)^5$$. 3. **Differentiate each part:** - For $$u(x)$$, use the chain rule: $$u'(x) = 7(2x + 5)^6 \cdot \frac{d}{dx}(2x + 5) = 7(2x + 5)^6 \cdot 2 = 14(2x + 5)^6$$. - For $$v(x)$$, also use the chain rule: $$v'(x) = 5(3x^4 - 8)^4 \cdot \frac{d}{dx}(3x^4 - 8) = 5(3x^4 - 8)^4 \cdot 12x^3 = 60x^3(3x^4 - 8)^4$$. 4. **Apply the product rule:** $$h'(x) = u'(x)v(x) + u(x)v'(x) = 14(2x + 5)^6 (3x^4 - 8)^5 + (2x + 5)^7 60x^3 (3x^4 - 8)^4$$. 5. **Factor common terms:** Both terms share $$ (2x + 5)^6 (3x^4 - 8)^4 $$, so factor it out: $$h'(x) = (2x + 5)^6 (3x^4 - 8)^4 \left[14(3x^4 - 8) + 60x^3 (2x + 5) \right]$$. 6. **Simplify inside the bracket:** $$14(3x^4 - 8) = 42x^4 - 112$$ $$60x^3 (2x + 5) = 120x^4 + 300x^3$$ Sum: $$42x^4 - 112 + 120x^4 + 300x^3 = (42x^4 + 120x^4) + 300x^3 - 112 = 162x^4 + 300x^3 - 112$$. 7. **Final derivative:** $$h'(x) = (2x + 5)^6 (3x^4 - 8)^4 (162x^4 + 300x^3 - 112)$$. This is the derivative of the given function using the product and chain rules.