What is the derivative of x^x?

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Answer 1 · 2018-03-05T01:24:29

$dy/dx=x^x(ln(x)+1)$

We have:

$y=x^x$ Let's take the natural log on both sides.

$ln(y)=ln(x^x)$ Using the fact that $log_a(b^c)=clog_a(b)$ ,

$=>ln(y)=xln(x)$ Apply $d/dx$ on both sides.

$=>d/dx(ln(y))=d/dx(xln(x))$

If $f(x)=g(h(x))$ , then $f'(x)=g'(h(x))*h'(x)$

$d/dx(x^n)=nx^(n-1)$ if $n$ is a constant.

Also, $d/dx(lnx)=1/x$

Lastly, the product rule:

If $f(x)=g(x)*h(x)$ , then $f'(x)=g'(x)*h(x)+g(x)*h'(x)$

We have:

$=>dy/dx*1/y=d/dx(x)*ln(x)+x*d/dx(ln(x))$

$=>dy/dx*1/y=1*ln(x)+x*1/x$

$=>dy/dx*1/y=ln(x)+cancelx*1/cancelx$
(Don't worry about when $x=0$ , because $ln(0)$ is undefined)

$=>dy/dx*1/y=ln(x)+1$

$=>dy/dx=y(ln(x)+1)$

Now, since $y=x^x$ , we can substitute $y$ .

$=>dy/dx=x^x(ln(x)+1)$