Question

y=1+lnx/x помогите пожалуйста​

Answer 1

$y'=\left(\frac{1+\ln x}{x}\right)'=\frac{(1+\ln x)'\times x-(1+\ln x)\times x'}{x^2}=\frac{\frac{1}{x}\times x-(1+\ln x)}{x^2}=-\frac{\ln x}{x^2}$

$y''=\left(-\frac{\ln x}{x^2}\right)'=-\frac{(\ln x)'\times x^2-\ln x\times (x^2)'}{x^4}=-\frac{\frac{1}{x}\times x^2-2x\ln x}{x^4}=\frac{2\ln x-1}{x^3}$