Question

Find the second order derivatives of the following :
$$x^{x}$$

Answer 1

$$y = x^{x}$$
$$\therefore \log y = \log x^{x} = x \log x$$
Differentiating both sides w.r.t. x, we get
$$\frac{1} {y} \cdot \frac{dy} {dx} = \frac{d} {dx} \left(x \log x\right)$$
$$\frac{1} {y} \cdot \frac{dy} {dx} = x \cdot \frac{d} {dx} \left( \log x\right) + \left( \log x\right) \frac{d} {dx} \left( x \right)$$
$$= \frac{x} {x} + \left( \log x\right) \left( 1\right)$$
$$= 1 + \log x$$
$$\therefore \frac{dy} {dx} = y \left(1 + \log x\right) = x^{x} \left(1 + \log x\right)$$
$$ \therefore \frac{d} {dx} \left(x^{x}\right) = x^{x} \left(1 + \log x\right)$$ ...........(1)
$$\therefore \frac{d^{2}y} {dx^{2}} = \frac{d} {dx} \left[x^{2} \left(1 + \log x\right) \right]$$
$$= x^{x} \cdot \frac{d} {dx} \left(1 + \log x\right) + \left(1 + \log x\right) \cdot \frac{d} {dx} \left(x^{x}\right)$$
$$ = x^{x} \left(0 + \frac{1} {x}\right) + \left(1 + \log x\right) \cdot x^{x} \left(1 + \log x\right)$$ .....[By (1)]
$$= x^{x-1} + x^{x} \left(1 + \log x\right)$$