Question Number 13688 by aishadenge last updated on 22/May/17
Answered by ajfour last updated on 22/May/17
$$\frac{\partial}{\partial{R}_{\mathrm{1}} }\left(\frac{\mathrm{1}}{{R}}\right)=−\frac{\mathrm{1}}{{R}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:−\frac{\mathrm{1}}{{R}^{\mathrm{2}} }\left(\frac{\partial{R}}{\partial{R}_{\mathrm{1}} }\right)=−\frac{\mathrm{1}}{{R}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\frac{\partial{R}}{\partial{R}_{\mathrm{1}} }=\frac{{R}^{\mathrm{2}} }{{R}_{\mathrm{1}} ^{\mathrm{2}} }\:\:\Rightarrow\:\:\:\frac{\partial^{\mathrm{2}} {R}}{\partial{R}_{\mathrm{1}} ^{\mathrm{2}} }=\frac{\mathrm{2}{RR}_{\mathrm{1}} ^{\mathrm{2}} \frac{\partial{R}}{\partial{R}_{\mathrm{1}} }−\mathrm{2}{R}_{\mathrm{1}} {R}^{\mathrm{2}} }{{R}_{\mathrm{1}} ^{\:\mathrm{4}} } \\ $$$$\:\:\:\:=\:\frac{\mathrm{2}{R}^{\mathrm{3}} −\mathrm{2}{R}_{\mathrm{1}} {R}^{\mathrm{2}} }{{R}_{\mathrm{1}} ^{\:\mathrm{4}} }=\frac{\mathrm{2}{R}^{\mathrm{3}} }{{R}_{\mathrm{1}} ^{\:\mathrm{4}} }\left(\mathrm{1}−\frac{{R}_{\mathrm{1}} }{{R}}\right) \\ $$$${so},\:\frac{\partial^{\mathrm{2}} \boldsymbol{{R}}}{\partial\boldsymbol{{R}}_{\mathrm{1}} }\:=\:\frac{\mathrm{2}}{\boldsymbol{{R}}_{\mathrm{1}} }\:\left(\frac{\boldsymbol{{R}}}{\boldsymbol{{R}}_{\mathrm{1}} }\right)^{\mathrm{3}} \left(\mathrm{1}−\frac{\boldsymbol{{R}}_{\mathrm{1}} }{\boldsymbol{{R}}}\right)\left(\partial\boldsymbol{{R}}_{\mathrm{1}} \right). \\ $$$$\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{2}}{{R}_{\mathrm{1}} }\left(\mathrm{1}−\frac{{R}}{{R}_{\mathrm{1}} }\right)\left(\frac{{R}}{{R}_{\mathrm{1}} }\right)^{\mathrm{2}} \left(\partial{R}_{\mathrm{1}} \right)\:\:\: \\ $$
Commented by aishadenge last updated on 22/May/17
$${thank}\:{you}. \\ $$