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Question Number 93478 by Ar Brandon last updated on 13/May/20

$$1\mathrm{\setminus }\mathrm{Calculate}{\mathrm{f}}_{\mathrm{x}}(2,3)\mathrm{if}\mathrm{f}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2+{\mathrm{y}}^2$$$$2\mathrm{\setminus }\mathrm{Calculate}\mathrm{df}(\mathrm{x},\mathrm{y})\mathrm{for}\mathrm{x}=1,\mathrm{y}=0,\mathrm{dx}=\frac{1}{2}$$$$\mathrm{and}\mathrm{dy}=\frac{1}{4}\mathrm{if}\mathrm{f}(\mathrm{x},\mathrm{y})=\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}$$

Commented by PRITHWISH SEN 2 last updated on 13/May/20

$$2.\mathbf{df}(\mathbf{x},\mathbf{y})=\left({\delta }_{\mathrm{x}}\mathrm{f}\right)\mathrm{dx}+\left({\delta }_{\mathrm{y}}\mathrm{f}\right)\mathrm{dy}=\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}\mathrm{dx}+\frac{\mathrm{y}}{\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}\mathrm{dy}$$$$=1.\frac{1}{2}+0.\frac{1}{4}=\frac{1}{2}\mathrm{please}\mathrm{check}$$$$$$

Commented by Ar Brandon last updated on 13/May/20

Thanks ��

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