How do you use logarithmic differentiation to find the derivative of the function $y = x^{x}$ $y=x^x$ ?

Question

3210 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-03-28T02:04:20

$y = x^{x}$ $y=x^x$

$ln y = {ln x}^{x}$ $lny=lnx^x$

$ln y = x ln x$ $lny=xlnx$

Now differentiate implicitly:

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

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

Solve for $\frac{d y}{d x}$ $(dy)/(dx)$

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

Recall that $y = x^{x}$ $y=x^x$ , so

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

Answer 2 · 2015-03-28T11:03:19

It's easy if you remember the logarithmic rule:

${[f (x)]}^{g (x)} = e^{ln ({[f (x)]}^{g (x)})} = e^{g (x) \cdot ln f (x)}$ $[f(x)]^g(x)=e^ln([f(x)]^g(x))=e^(g(x)*lnf(x)$ ,

so:

$y = x^{x} = e^{x ln x} \Rightarrow$ $y=x^x=e^(xlnx)rArr$

$y'=e^(xlnx)(1*lnx+x*1/x)=e^(xlnx)(lnx+1)$ ,

(or, if you want: $y'=x^x(lnx+1)$ ).

How do you use logarithmic differentiation to find the derivative of the function y=x^xy=xxy=x^x?