(uv) =u v+uv The product changes because each factor changes. Both first-order contributions must be included. Method Choose the two factors; differentiate each in turn while holding the other unchanged; add; factor if useful. Example y=xe2x, y =e2x+2xe2x=e2x(1+2x) Exam check Multiplying the two derivatives loses both first-order contributions. One factor being zero does not force the product derivative to be zero.

\((uv)^{\prime}=u^{\prime}v+uv^{\prime}\)

The product changes because each factor changes. Both first-order contributions must be included.

Choose the two factors; differentiate each in turn while holding the other unchanged; add; factor if useful.

\(y=xe^{2x},\quad y^{\prime}=e^{2x}+2xe^{2x}=e^{2x}(1+2x)\)

Multiplying the two derivatives loses both first-order contributions. One factor being zero does not force the product derivative to be zero.

Products of functions

Products of functions