## Calculus 8th Edition

$F'(x) = 24x^{11}(5x^3 + 2)(5x^3+1)$
Original Expression: $F(x) = (5x^6 + 2x^3)^4$ $u = 5x^6 + 2x^3$ $F(u) = x^4$ Apply the chain rule: $f'(u) \times u'$ $F'(u) = 4(u)^3 30x^5 +6x^2$ $F'(x) = 4(5x^6 + 2x^3)^3 (30x^5 +6x^2)$ Simplify: $4(x^3)^3(5x^3 + 2)6x^2(5x^3+1)$ $4x^9(5x^3 + 2)6x^2(5x^3+1)$ $24x^{11}(5x^3 + 2)(5x^3+1)$