Question Number 26297 by d.monhbayr@gmail.com last updated on 23/Dec/17
$${y}={x}−\frac{\mathrm{2}}{{x}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} } \\ $$$${y}'=? \\ $$
Answered by Joel578 last updated on 24/Dec/17
$$\:{y}\:=\:{x}\:−\:\mathrm{2}{x}^{−\mathrm{4}} \:−\:\frac{\mathrm{1}}{\mathrm{3}}{x}^{−\mathrm{3}} \\ $$$${y}'\:=\:\mathrm{1}\:+\:\mathrm{8}{x}^{−\mathrm{5}} \:+\:{x}^{−\mathrm{4}} \\ $$