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\[\ln x \sqrt{x+1} =y\]

rewrite first as
\(y=\ln(x)+\frac{1}{2}\ln(x+1)\)
Therefore
\(y'=\frac{1}{x}+\frac{1}{2(x+1)}\)

\[\ln (x \sqrt{x+1}) =y\] ?

@BlingBlong is it the way watchmath has shown
thanks@ watchmath!