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We need to differentiate y = log(x³ - x) + 7(2/3)x. We'll differentiate each term separately.First term: y₁ = log(x³ - x)Assuming log is the natural logarithm (ln):dy₁/dx = d/dx(ln(x³ - x))Using the chain rule, d/dx(ln(u)) = u'/u:dy₁/dx = (d/dx(x³ - x)) / (x³ - x) = (3x² - 1) / (x³ - x)Since x³ - x = x(x² - 1), we have dy₁/dx = (3x² - 1) / (x(x² - 1)).Second term: y₂ = 7(2/3)xThis is of the form au, where a=7 and u=(2/3)x. The derivative is d/dx(au) = au ln(a) (du/dx).dy₂/dx = 7(2/3)x ln(7) (d/dx((2/3)x))dy₂/dx = 7(2/3)x ln(7) (2/3)Since 'log7' is used in the options, we assume ln(7) is represented as log7.dy₂/dx = 7(2/3)x (2/3 log7).Total derivative:dy/dx = dy₁/dx + dy₂/dx = (3x² - 1) / (x(x² - 1)) + 7(2/3)x (2/3 log7).This matches option D.