MCQ
$\frac{d}{{dx}}\left( {{x^x}} \right) = ........\left( {x > 0} \right)$
- A${x^{x - 1}}$
- B$x^x$
- C$0$
- ✓${x^x}\left( {1 + \log x} \right)$
$\frac{d}{{dx}}{x^x} = \frac{d}{{dx}}{e^{x\,\,\log x}}$
$ = {e^{x\,\log x}}\frac{d}{{dx}}x\log x$
$ = {x^x}\left( {x\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}x} \right)$
$ = {x^x}\left( {\frac{x}{x} + \log x} \right) = {x^x}\left( {1 + \log x} \right)$
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