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To differentiate y = log(x3 - x) + 7(2/3)x, we differentiate each term separately.Assume 'log' means natural logarithm (ln).For the first term, d/dx(ln(x3 - x)):Using the chain rule, d/dx(ln u) = (1/u) * du/dx.= (1 / (x3 - x)) * d/dx(x3 - x)= (1 / (x3 - x)) * (3x2 - 1)= (3x2 - 1) / (x(x2 - 1)).For the second term, d/dx(7(2/3)x):Using the rule d/dx(au) = au * ln a * du/dx, where a = 7 and u = (2/3)x.du/dx = 2/3.= 7(2/3)x * ln 7 * (2/3).Combine the derivatives of both terms:dy/dx = (3x2 - 1) / (x(x2 - 1)) + (2/3) * 7(2/3)x * ln 7.The term 'log7' in options typically refers to 'ln 7'. Thus option D is correct.