Question

f(x) = $x^{x}$ найти производную

Answer 1

Ответ:

$f(x) = {x}^{x}$

$f'(x) = \frac{d}{dx} ( {x}^{x} )$

$f'(x) = \frac{d}{dx} ({( {e}^{ln(x)} )^{x}} )$

$f'(x) = \frac{d}{dx} ( {e}^{ln(x) \times x} )$

$f'(x) = \frac{d}{dg} ( {e}^{g} ) \times \frac{d}{dx} (ln(x) \times x)$

$\frac{d}{dg} ( {e}^{g} ) \\ {e}^{g}$

$\frac{d}{dx} (ln(x) \times x) \\ \frac{d}{dx} (ln(x)) \times x + ln(x) \times \frac{d}{dx} (x) \\ \frac{1}{x} \times x + ln(x) \times 1 \\ \frac{1}{x} \times x + ln(x)$

$f'(x) = {e}^{g} \times ( \frac{1}{x} \times x + ln(x))$

$f'(x) = {e}^{ln(x) \times x} \times ( \frac{1}{x} \times x + ln(x))$

$f'(x) = {e}^{x \times ln(x)} \times (1 + ln(x))$

$f'(x) = {e}^{ln( {x}^{x}) } \times (1 + ln(x))$

$f'(x) = {x}^{x} \times (1 + ln(x))$

$f'(x) = {x}^{x} + {x}^{x} \times ln(x)$