Question Number 96600 by Hamida last updated on 03/Jun/20
Answered by Rio Michael last updated on 03/Jun/20
$$\:{y}\:=\:\frac{\mathrm{4}}{\mathrm{7}}{x}^{\mathrm{3}} \:−\:\frac{\mathrm{2}}{\mathrm{3}}{x}^{−\mathrm{1}} \\ $$$${y}\:'\:=\:\frac{\mathrm{12}}{\mathrm{7}}{x}^{\mathrm{3}} \:+\:\frac{\mathrm{2}}{\mathrm{3}{x}^{\mathrm{2}} }\: \\ $$$${y}'\mid_{{x}=\mathrm{7}} \:=\:\mathrm{12}\left(\mathrm{7}\right)^{\mathrm{2}} \:+\:\frac{\mathrm{2}}{\mathrm{3}\left(\mathrm{7}^{\mathrm{2}} \right)} \\ $$