Answer
$y'=\frac{1}{x(1+lnx)}$ and $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$
Work Step by Step
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 $y''=-\frac{2+lnx}{x^{2}(1+lnx)^{2}}$
