Question Number 37662 by Rio Mike last updated on 16/Jun/18
$$\:\mathrm{Given}\:\mathrm{that}\:{y}={x}^{\mathrm{2}} {cosx}, \\ $$$$\mathrm{find}\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}\:},\:\mathrm{simplifying}\:\mathrm{your}\:\mathrm{answer} \\ $$$$\mathrm{as}\:\mathrm{far}\:\mathrm{as}\:\mathrm{posible} \\ $$
Answered by Joel579 last updated on 16/Jun/18
$${u}\:=\:{x}^{\mathrm{2}} \\ $$$${v}\:=\:\mathrm{cos}\:{x} \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}\:=\:{uv}'\:+\:{vu}'\:=\:{x}^{\mathrm{2}} \:.\:−\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}\:.\:\mathrm{2}{x} \\ $$$$\:\:\:\:\:\:\:=\:−{x}^{\mathrm{2}} \:\mathrm{sin}\:{x}\:+\:\mathrm{2}{x}\:\mathrm{cos}\:{x} \\ $$$$\:\:\:\:\:\:\:=\:{x}\left(\mathrm{2cos}\:{x}\:−\:{x}\:\mathrm{sin}\:{x}\right) \\ $$