Question Detail
Question
Differentiate y = log(x3 - x) + 7(3x/2) with respect to x.
Correct Answer
Option D is the correct answer.
Detailed Explanation
To differentiate y = log(x3 - x) + 7(3x/2), we differentiate each term separately.For the first term, log(x3 - x), assuming 'log' refers to the natural logarithm (ln):d/dx [ln(x3 - x)] = (1 / (x3 - x)) * d/dx (x3 - x) = (3x2 - 1) / (x3 - x) = (3x2 - 1) / (x(x2 - 1)).For the second term, 7(3x/2), using the rule d/dx (au) = au * ln(a) * u':u = 3x/2, so u' = 3/2. Also a = 7.d/dx [7(3x/2)] = 7(3x/2) * ln(7) * (3/2).Combining these, dy/dx = (3x2 - 1) / (x(x2 - 1)) + (3/2)ln(7) * 7(3x/2).Option D matches the first term exactly. The second term in option D, 7(3x/2) (2/3 log7), appears to have a typo where the coefficient 3/2 for ln(7) is written as 2/3 log7. However, assuming 'log' implies 'ln', and matching the structure, option D is the intended answer.
Hint
Remember the chain rule for logarithmic functions: d/dx [ln(f(x))] = f'(x)/f(x), and the differentiation rule for exponential functions with a constant base: d/dx [ag(x)] = ag(x) * ln(a) * g'(x).