Question Number 224735 by fantastic last updated on 30/Sep/25
$${Let}\:{u}=\frac{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} {y}^{\mathrm{2}} },\:{v}=\frac{{z}^{\mathrm{2}} −{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} {z}^{\mathrm{2}} }\:{for}\:{x}\neq\mathrm{0},{y}\neq\mathrm{0}{z}\neq\mathrm{0}. \\ $$$${Let}\:{w}={f}\left({u},{v}\right),\:{where}\:{f}\:{is}\:{a}\:{real} \\ $$$${valued}\:{function}\:{defined}\:{on}\:{R}^{\mathrm{2}} \\ $$$${having}\:{continuous}\:{first}\:{order} \\ $$$${partial}\:{derivatives}. \\ $$$${the}\:{value}\:{of} \\ $$$${x}^{\mathrm{3}} \:\frac{\partial{w}}{\partial{x}}+{y}^{\mathrm{3}} \:\frac{\partial{w}}{\partial{y}}+{z}^{\mathrm{3}} \:\frac{\partial{w}}{\partial{z}}\:{at}\:{point}\:\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:{is} \\ $$
Answered by mr W last updated on 04/Oct/25
$${u}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\partial{u}}{\partial{x}}=−\frac{\mathrm{2}}{{x}^{\mathrm{3}} },\:\frac{\partial{u}}{\partial{y}}=\frac{\mathrm{2}}{{y}^{\mathrm{3}} },\:\frac{\partial{u}}{\partial{z}}=\mathrm{0} \\ $$$${v}=\frac{\mathrm{1}}{{y}^{\mathrm{2}} }−\frac{\mathrm{1}}{{z}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{\partial{v}}{\partial{x}}=\mathrm{0},\:\frac{\partial{v}}{\partial{y}}=−\frac{\mathrm{2}}{{y}^{\mathrm{3}} },\:\frac{\partial{v}}{\partial{z}}=\frac{\mathrm{2}}{{z}^{\mathrm{3}} } \\ $$$$\frac{\partial{w}}{\partial{x}}=\frac{\partial{w}}{\partial{u}}×\frac{\partial{u}}{\partial{x}}+\frac{\partial{w}}{\partial{v}}×\frac{\partial{v}}{\partial{x}}=−\frac{\mathrm{2}}{{x}^{\mathrm{3}} }×\frac{\partial{w}}{\partial{u}}+\frac{\partial{w}}{\partial{v}}×\mathrm{0} \\ $$$$\frac{\partial{w}}{\partial{y}}=\frac{\partial{w}}{\partial{u}}×\frac{\partial{u}}{\partial{y}}+\frac{\partial{w}}{\partial{v}}×\frac{\partial{v}}{\partial{y}}=\frac{\mathrm{2}}{{y}^{\mathrm{3}} }×\frac{\partial{w}}{\partial{u}}−\frac{\mathrm{2}}{{y}^{\mathrm{3}} }×\frac{\partial{w}}{\partial{v}} \\ $$$$\frac{\partial{w}}{\partial{z}}=\frac{\partial{w}}{\partial{u}}×\frac{\partial{u}}{\partial{z}}+\frac{\partial{w}}{\partial{v}}×\frac{\partial{v}}{\partial{z}}=\frac{\partial{w}}{\partial{u}}×\mathrm{0}+\frac{\mathrm{2}}{{z}^{\mathrm{3}} }×\frac{\partial{w}}{\partial{v}} \\ $$$${x}^{\mathrm{3}} \frac{\partial{w}}{\partial{x}}+{y}^{\mathrm{3}} \frac{\partial{w}}{\partial{y}}+{z}^{\mathrm{3}} \frac{\partial{w}}{\partial{z}} \\ $$$$={x}^{\mathrm{3}} \left(−\frac{\mathrm{2}}{{x}^{\mathrm{3}} }×\frac{\partial{w}}{\partial{u}}\right)+{y}^{\mathrm{3}} \left(\frac{\mathrm{2}}{{y}^{\mathrm{3}} }×\frac{\partial{w}}{\partial{u}}−\frac{\mathrm{2}}{{y}^{\mathrm{3}} }×\frac{\partial{w}}{\partial{v}}\right)+{z}^{\mathrm{3}} \left(\frac{\mathrm{2}}{{z}^{\mathrm{3}} }×\frac{\partial{w}}{\partial{v}}\right) \\ $$$$=−\mathrm{2}\frac{\partial{w}}{\partial{u}}+\mathrm{2}\frac{\partial{w}}{\partial{u}}−\mathrm{2}\frac{\partial{w}}{\partial{v}}+\mathrm{2}\frac{\partial{w}}{\partial{v}} \\ $$$$=\mathrm{0}\:\checkmark \\ $$