1. **Problem statement:** Find the derivative of the function $$f(x) = \frac{\sqrt{x-1} - \frac{5}{\sqrt{x-1}}}{\sqrt[3]{x-1}}$$ and then evaluate $f'(2)$. 2. **Formula and rules:** We will use the quotient rule for derivatives: $$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ where $u = \sqrt{x-1} - \frac{5}{\sqrt{x-1}}$ and $v = \sqrt[3]{x-1} = (x-1)^{1/3}$. Also, recall the derivatives: - $\frac{d}{dx} (x-1)^{m} = m (x-1)^{m-1}$ - $\frac{d}{dx} \sqrt{x-1} = \frac{1}{2\sqrt{x-1}}$ - $\frac{d}{dx} \frac{1}{\sqrt{x-1}} = -\frac{1}{2(x-1)^{3/2}}$ 3. **Calculate $u$ and $u'$:** $$u = (x-1)^{1/2} - 5 (x-1)^{-1/2}$$ $$u' = \frac{1}{2} (x-1)^{-1/2} - 5 \left(-\frac{1}{2} (x-1)^{-3/2}\right) = \frac{1}{2} (x-1)^{-1/2} + \frac{5}{2} (x-1)^{-3/2}$$ 4. **Calculate $v$ and $v'$:** $$v = (x-1)^{1/3}$$ $$v' = \frac{1}{3} (x-1)^{-2/3}$$ 5. **Apply quotient rule:** $$f'(x) = \frac{u'v - uv'}{v^2}$$ Substitute: $$= \frac{\left(\frac{1}{2} (x-1)^{-1/2} + \frac{5}{2} (x-1)^{-3/2}\right)(x-1)^{1/3} - \left((x-1)^{1/2} - 5 (x-1)^{-1/2}\right) \frac{1}{3} (x-1)^{-2/3}}{\left((x-1)^{1/3}\right)^2}$$ 6. **Simplify powers of $(x-1)$:** - Multiply terms: $$\left(\frac{1}{2} (x-1)^{-1/2} + \frac{5}{2} (x-1)^{-3/2}\right)(x-1)^{1/3} = \frac{1}{2} (x-1)^{-1/2 + 1/3} + \frac{5}{2} (x-1)^{-3/2 + 1/3}$$ Calculate exponents: $$-\frac{1}{2} + \frac{1}{3} = -\frac{3}{6} + \frac{2}{6} = -\frac{1}{6}$$ $$-\frac{3}{2} + \frac{1}{3} = -\frac{9}{6} + \frac{2}{6} = -\frac{7}{6}$$ So: $$= \frac{1}{2} (x-1)^{-1/6} + \frac{5}{2} (x-1)^{-7/6}$$ - For the second term: $$\left((x-1)^{1/2} - 5 (x-1)^{-1/2}\right) \frac{1}{3} (x-1)^{-2/3} = \frac{1}{3} (x-1)^{1/2 - 2/3} - \frac{5}{3} (x-1)^{-1/2 - 2/3}$$ Calculate exponents: $$\frac{1}{2} - \frac{2}{3} = \frac{3}{6} - \frac{4}{6} = -\frac{1}{6}$$ $$-\frac{1}{2} - \frac{2}{3} = -\frac{3}{6} - \frac{4}{6} = -\frac{7}{6}$$ So: $$= \frac{1}{3} (x-1)^{-1/6} - \frac{5}{3} (x-1)^{-7/6}$$ 7. **Combine numerator:** $$\left(\frac{1}{2} (x-1)^{-1/6} + \frac{5}{2} (x-1)^{-7/6}\right) - \left(\frac{1}{3} (x-1)^{-1/6} - \frac{5}{3} (x-1)^{-7/6}\right)$$ $$= \left(\frac{1}{2} - \frac{1}{3}\right) (x-1)^{-1/6} + \left(\frac{5}{2} + \frac{5}{3}\right) (x-1)^{-7/6}$$ Calculate coefficients: $$\frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6}$$ $$\frac{5}{2} + \frac{5}{3} = \frac{15}{6} + \frac{10}{6} = \frac{25}{6}$$ So numerator is: $$\frac{1}{6} (x-1)^{-1/6} + \frac{25}{6} (x-1)^{-7/6}$$ 8. **Denominator:** $$\left((x-1)^{1/3}\right)^2 = (x-1)^{2/3}$$ 9. **Final derivative expression:** $$f'(x) = \frac{\frac{1}{6} (x-1)^{-1/6} + \frac{25}{6} (x-1)^{-7/6}}{(x-1)^{2/3}} = \frac{1}{6} \frac{(x-1)^{-1/6} + 25 (x-1)^{-7/6}}{(x-1)^{2/3}}$$ Rewrite denominator as negative exponent: $$= \frac{1}{6} \left((x-1)^{-1/6 - 2/3} + 25 (x-1)^{-7/6 - 2/3}\right)$$ Calculate exponents: $$-\frac{1}{6} - \frac{2}{3} = -\frac{1}{6} - \frac{4}{6} = -\frac{5}{6}$$ $$-\frac{7}{6} - \frac{2}{3} = -\frac{7}{6} - \frac{4}{6} = -\frac{11}{6}$$ So: $$f'(x) = \frac{1}{6} \left((x-1)^{-5/6} + 25 (x-1)^{-11/6}\right)$$ 10. **Evaluate at $x=2$:** Calculate powers: $$2 - 1 = 1$$ $$1^{-5/6} = 1$$ $$1^{-11/6} = 1$$ So: $$f'(2) = \frac{1}{6} (1 + 25) = \frac{26}{6} = \frac{13}{3}$$ **Final answer:** $$f'(2) = \frac{13}{3}$$