Question #86860

Question

1116 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-01T09:15:56

$f'(x)=-(ln(x)+1)/x^x$

$f(x)=(1/x)^x$

$f'(x)=d/dx[(1/x)^x]=d/dx[1^x/x^x]=d/dx[1/x^x]$

$color(white)(f'(x))=(x^xd/dx[1]-1d/dx[x^x])/(x^x)^2$

$color(white)(f'(x))=(x^x(0)-1(x^xd/dx[xln(x)]))/x^(2x)$

$color(white)(f'(x))=-(x^xd/dx[xln(x)])/x^(2x)$

$color(white)(f'(x))=-(x^x(ln(x)d/dx[x]+xd/dx[ln(x)]))/x^(2x)$

$color(white)(f'(x))=-(x^x(ln(x)(1)+x(1/x)))/x^(2x)$

$color(white)(f'(x))=-(x^x(ln(x)+1))/x^(2x)$

$color(white)(f'(x))=-(ln(x)+1)/x^x$