Question Number 70583 by Raphael last updated on 05/Oct/19
$$\mathrm{Given}\:\mathrm{that}\:\mathrm{y}=\frac{\mathrm{4}}{\:\sqrt{\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)}}\:\mathrm{show}\:\mathrm{that}\:\mathrm{2}\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=−\mathrm{3x}^{\mathrm{2}} \mathrm{y}. \\ $$
Answered by MJS last updated on 05/Oct/19
$${dy}=−\frac{\mathrm{6}{x}^{\mathrm{2}} }{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }{dx} \\ $$$$\mathrm{2}\left({x}^{\mathrm{3}} +\mathrm{1}\right)\frac{{dy}}{{dc}}=−\frac{\mathrm{12}{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}}=−\mathrm{3}{x}^{\mathrm{2}} \frac{\mathrm{4}}{\:\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}}=−\mathrm{3}{x}^{\mathrm{2}} {y} \\ $$