Question

If \[ f(x) = \sqrt[3]{x^2} + \sqrt{x}, \] then the value of \( f'(64) \) is equal to

• A. \( \frac{11}{48} \)
• B. \( \frac{9}{48} \)
• C. \( \frac{7}{48} \)
• D. \( \frac{5}{48} \)
• E. \( \frac{1}{16} \)

Hint:

To differentiate power functions, use \( \frac{d}{dx} x^n = n x^{n-1} \).

Answer 1

The Correct Option is A

Differentiating \( f(x) \), \[ f(x) = x^{\frac{2}{3}} + x^{\frac{1}{2}}. \] Applying differentiation, \[ f'(x) = \frac{2}{3} x^{-\frac{1}{3}} + \frac{1}{2} x^{-\frac{1}{2}}. \] Evaluating at \( x = 64 \), \[ f'(64) = \frac{2}{3} \cdot 64^{-\frac{1}{3}} + \frac{1}{2} \cdot 64^{-\frac{1}{2}}. \] Since \( 64^{\frac{1}{3}} = 4 \) and \( 64^{\frac{1}{2}} = 8 \), \[ f'(64) = \frac{2}{3} \times \frac{1}{4} + \frac{1}{2} \times \frac{1}{8}. \] \[ = \frac{2}{12} + \frac{1}{16}. \] \[ = \frac{1}{6} + \frac{1}{16}. \] \[ = \frac{8}{48} + \frac{3}{48} = \frac{11}{48}. \] Thus, the correct answer is (A).