Answer is: option1
\( -\frac{1}{20} \)Solution:
\[ y=x\ln x \]
\[ y'=\ln x + x\left(\frac{1}{x}\right)=\ln x+1 \]
\[ y''=\frac{1}{x},\qquad y'''=-\frac{1}{x^2} \]
Continuing: \[ y^{(4)}=\frac{2}{x^3},\qquad y^{(5)}=-\frac{6}{x^4} \]
\[ y^{(5)}(1)=-6 \]
The desired coefficient is \[ \frac{y^{(5)}(1)}{5!} = \frac{-6}{120} = -\frac{1}{20} \]
