Question

If y(x) = xx, x > 0, then y''(2) - 2y'(2) is equal to

Answer 1

4(ln2)^2 - 2

Answer 2

Solution Explanation:

1. Write $y(x) = x^x = e^{x\ln x}$.
2. Differentiate to find $y'(x)$: $$y'(x) = e^{x\ln x} \cdot \frac{d}{dx}(x\ln x) = x^x(1+\ln x)$$
3. Differentiate to find $y''(x)$: $$y''(x) = \frac{d}{dx}\left[x^x(1+\ln x)\right] = x^x(1+\ln x)^2 + x^x\cdot\frac{1}{x} = x^x\left[(1+\ln x)^2+\frac{1}{x}\right]$$
4. Substitute $x = 2$: $$y'(2) = 2^2\,(1+\ln 2) = 4(1+\ln 2)$$ $$y''(2) = 2^2\left[(1+\ln2)^2+\frac{1}{2}\right] = 4\left[(1+\ln2)^2+\frac{1}{2}\right]$$
5. Calculate $y''(2) - 2y'(2)$: $$y''(2) - 2y'(2) = 4\left[(1+\ln2)^2+\frac{1}{2}\right] - 8(1+\ln2)$$ Expand and simplify: $$= 4(1+\ln2)^2 + 2 - 8(1+\ln2)$$ $$= 4\left[(\ln2)^2 + 2\ln2 + 1\right] + 2 - 8\ln2 - 8$$ $$= 4(\ln2)^2 + 8\ln2 + 4 + 2 - 8\ln2 - 8$$ $$= 4(\ln2)^2 - 2$$