Denis virtually did

[tex]f'(x) = x^2[6(2x + 1)^2] + (2x + 1)^3(2x) = 6x^2k^2 + 2xk^3, \text { where } k = 2x + 1.[/tex]

[tex]\therefore f'(x) = 6x^2k^2 + 2xk^3 = 2(3x^2k^2 + xk^3) = 2x(3xk^2 + k^3) = 2xk^2(3x + k) =[/tex]

[tex]2xk^2(3x + 2x + 1) = 2xk^2(5x + 1) = 2x(2x + 1)^2(5x + 1) \ \because \ k = 2x + 1.[/tex]

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