Question #86860

1 Answer
Feb 1, 2018

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

Explanation:

#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#