## Calculus: Early Transcendentals 8th Edition

$y′=x(2sin(πx)+πxcos(πx))$
Start with the function: $y=x^2sin(πx)$. Use the product rule to differentiate: $y′=2xsin(πx)+πx^2cos(πx)$. Factor out the common $x$ term to arrive at the answer: $y′=x(2sin(πx)+πxcos(πx))$.