Let's perform the differentiation: (x·ln(x))' We differentiate U1 = x and V2 = ln(x) using: (U1·V2)' = (U1)'·V2 + U1·(V2)' Let's perform the differentiation U1: (x)' 1 Let's differentiate V2: (ln(x))' 1x You can evaluate the resulted derivative (U1·V2)': 1·ln(x) + x·1x ▷Answer: ln(x) + 1