## Calculus: Early Transcendentals 8th Edition

$y'=\frac{1}{x(1+lnx)}$ and $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$
Find the first and second derivative of the function $y=ln(1+lnx)$ $y'=\frac{1}{x(1+lnx)}$ and $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$ Differentiate the given function with respect to $x$. $y=\frac{d}{dx}ln(1+lnx)$ Thus, $y'=\frac{1}{x(1+lnx)}$ Now, $y''=\frac{d}{dx}\frac{1}{x(1+lnx)}$ $=\frac{1}{x^{2}(1+lnx)^{2}}\frac{d}{dx}\frac{1}{x(1+lnx)}$ $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$ Hence, $y'=\frac{1}{x(1+lnx)}$ and $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$