1. **State the problem:** Find the third derivative $y^{(3)}$ of the function $$y = \frac{\lg(x - 1)}{x^3}$$ where $\lg$ denotes the base-10 logarithm. 2. **Recall the formula and rules:** - The derivative of $\lg u$ with respect to $x$ is $$\frac{1}{u \ln 10} \cdot \frac{du}{dx}$$. - Use the quotient rule for derivatives: $$\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$$. - Higher order derivatives are found by differentiating repeatedly. 3. **Rewrite the function:** $$y = \frac{\lg(x-1)}{x^3}$$ Let $f(x) = \lg(x-1)$ and $g(x) = x^3$. 4. **First derivative $y'$:** $$f'(x) = \frac{1}{(x-1) \ln 10}$$ $$g'(x) = 3x^2$$ Using quotient rule: $$y' = \frac{f'g - fg'}{g^2} = \frac{\frac{1}{(x-1) \ln 10} \cdot x^3 - \lg(x-1) \cdot 3x^2}{x^6}$$ Simplify numerator: $$= \frac{x^3}{(x-1) \ln 10} - 3x^2 \lg(x-1)$$ So, $$y' = \frac{\frac{x^3}{(x-1) \ln 10} - 3x^2 \lg(x-1)}{x^6} = \frac{x^3}{(x-1) \ln 10 \cdot x^6} - \frac{3x^2 \lg(x-1)}{x^6} = \frac{1}{(x-1) \ln 10 \cdot x^3} - \frac{3 \lg(x-1)}{x^4}$$ 5. **Second derivative $y''$:** Differentiate each term separately: - For $$\frac{1}{(x-1) \ln 10 \cdot x^3} = \frac{1}{\ln 10} \cdot \frac{1}{(x-1) x^3}$$ Use product rule on $$\frac{1}{(x-1) x^3} = (x-1)^{-1} x^{-3}$$: $$\frac{d}{dx}[(x-1)^{-1} x^{-3}] = (x-1)^{-1} \cdot (-3x^{-4}) + x^{-3} \cdot (-1)(x-1)^{-2} = -3 \frac{1}{(x-1) x^4} - \frac{1}{(x-1)^2 x^3}$$ Multiply by $$\frac{1}{\ln 10}$$: $$- \frac{3}{(x-1) x^4 \ln 10} - \frac{1}{(x-1)^2 x^3 \ln 10}$$ - For $$- \frac{3 \lg(x-1)}{x^4}$$ use product rule: $$u = \lg(x-1), \quad v = x^{-4}$$ $$u' = \frac{1}{(x-1) \ln 10}, \quad v' = -4x^{-5}$$ So, $$\frac{d}{dx}[-3 u v] = -3 (u' v + u v') = -3 \left( \frac{1}{(x-1) \ln 10} x^{-4} + \lg(x-1)(-4x^{-5}) \right)$$ $$= -3 \frac{1}{(x-1) \ln 10 x^4} + 12 \frac{\lg(x-1)}{x^5}$$ 6. **Combine second derivative terms:** $$y'' = - \frac{3}{(x-1) x^4 \ln 10} - \frac{1}{(x-1)^2 x^3 \ln 10} - 3 \frac{1}{(x-1) x^4 \ln 10} + 12 \frac{\lg(x-1)}{x^5}$$ Combine like terms: $$y'' = - \frac{6}{(x-1) x^4 \ln 10} - \frac{1}{(x-1)^2 x^3 \ln 10} + 12 \frac{\lg(x-1)}{x^5}$$ 7. **Third derivative $y^{(3)}$:** Differentiate each term: - For $$- \frac{6}{(x-1) x^4 \ln 10} = - \frac{6}{\ln 10} (x-1)^{-1} x^{-4}$$ Derivative: $$- \frac{6}{\ln 10} \left( (x-1)^{-1} (-4x^{-5}) + x^{-4} (-1)(x-1)^{-2} \right) = - \frac{6}{\ln 10} \left( -4 \frac{1}{(x-1) x^5} - \frac{1}{(x-1)^2 x^4} \right)$$ $$= \frac{24}{(x-1) x^5 \ln 10} + \frac{6}{(x-1)^2 x^4 \ln 10}$$ - For $$- \frac{1}{(x-1)^2 x^3 \ln 10} = - \frac{1}{\ln 10} (x-1)^{-2} x^{-3}$$ Derivative: $$- \frac{1}{\ln 10} \left( -2 (x-1)^{-3} x^{-3} + (x-1)^{-2} (-3 x^{-4}) \right) = - \frac{1}{\ln 10} \left( -2 \frac{1}{(x-1)^3 x^3} - 3 \frac{1}{(x-1)^2 x^4} \right)$$ $$= \frac{2}{(x-1)^3 x^3 \ln 10} + \frac{3}{(x-1)^2 x^4 \ln 10}$$ - For $$12 \frac{\lg(x-1)}{x^5} = 12 \lg(x-1) x^{-5}$$ Derivative: $$12 \left( \frac{1}{(x-1) \ln 10} x^{-5} + \lg(x-1)(-5 x^{-6}) \right) = \frac{12}{(x-1) x^5 \ln 10} - 60 \frac{\lg(x-1)}{x^6}$$ 8. **Combine all terms for $y^{(3)}$:** $$y^{(3)} = \frac{24}{(x-1) x^5 \ln 10} + \frac{6}{(x-1)^2 x^4 \ln 10} + \frac{2}{(x-1)^3 x^3 \ln 10} + \frac{3}{(x-1)^2 x^4 \ln 10} + \frac{12}{(x-1) x^5 \ln 10} - 60 \frac{\lg(x-1)}{x^6}$$ Group like terms: $$y^{(3)} = \left( \frac{24 + 12}{(x-1) x^5 \ln 10} \right) + \left( \frac{6 + 3}{(x-1)^2 x^4 \ln 10} \right) + \frac{2}{(x-1)^3 x^3 \ln 10} - 60 \frac{\lg(x-1)}{x^6}$$ $$= \frac{36}{(x-1) x^5 \ln 10} + \frac{9}{(x-1)^2 x^4 \ln 10} + \frac{2}{(x-1)^3 x^3 \ln 10} - 60 \frac{\lg(x-1)}{x^6}$$ **Final answer:** $$\boxed{y^{(3)} = \frac{36}{(x-1) x^5 \ln 10} + \frac{9}{(x-1)^2 x^4 \ln 10} + \frac{2}{(x-1)^3 x^3 \ln 10} - 60 \frac{\lg(x-1)}{x^6}}$$