Differentiate $y = x^{x}$ $y=x^x$ ?

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Answer 1 · 2016-03-27T07:47:45

$\frac{d y}{d x} = (1 + ln x) x^{x}$ $(dy)/(dx)=(1+lnx)x^x$

As $y = x^{x}$ $y=x^x$ , we have $ln y = {ln x}^{x} = x ln x$ $lny=lnx^x=xlnx$

Hence differentiating both sides, $\frac{1}{y} \times \frac{d y}{d x} = 1 \times ln x + x \times \frac{1}{x}$ $1/yxx(dy)/(dx)=1xxlnx+x xx1/x$

or $\frac{1}{x^{x}} \times \frac{d y}{d x} = 1 + ln x$ $1/(x^x)xx(dy)/(dx)=1+lnx$ and hence

$\frac{d y}{d x} = (1 + ln x) x^{x}$ $(dy)/(dx)=(1+lnx)x^x$

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