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Question Number 88434 by Rio Michael last updated on 10/Apr/20

$$\mathrm{find}\mathrm{the}\mathrm{partial}\mathrm{derivatives}\mathrm{of}\mathrm{first}\mathrm{and}\mathrm{second}\mathrm{order}$$$$\mathrm{for}\mathrm{the}\mathrm{function}$$$$f(x,y)=x^{3}y+3xy+y^4$$

Commented by niroj last updated on 10/Apr/20

$$\mathrm{f}(\mathrm{x},\mathrm{y})={\mathrm{x}}^3\mathrm{y}+3\mathrm{xy}+{\mathrm{y}}^4$$$$\mathrm{D}.\mathrm{w}.\mathrm{r}.\mathrm{to}\mathrm{x}$$$$1^{\mathrm{st}}\mathrm{order}\mathrm{f}\left({\mathrm{x}}^{{}^{\prime }}\right)=\frac{\mathrm{df}}{\mathrm{dx}}={3\mathrm{x}}^2\mathrm{y}+3\mathrm{y}$$$$2^{\mathrm{nd}}\mathrm{order}\mathrm{f}\left({\mathrm{x}}^{{}^{″}}\right)=\frac{{\mathrm{d}}^2\mathrm{f}}{{\mathrm{dx}}^2}=6\mathrm{xy}.$$$$\mathrm{Again},\mathrm{f}\left(\mathrm{y}\right)={\mathrm{x}}^3\mathrm{y}+3\mathrm{xy}+{\mathrm{y}}^4$$$$\mathrm{D}.\mathrm{w}.\mathrm{r}.\mathrm{to}\mathrm{y}$$$$1^{\mathrm{st}}\mathrm{order}\mathrm{f}\left({\mathrm{y}}^{{}^{\prime }}\right)=\frac{\mathrm{df}}{\mathrm{dy}}={\mathrm{x}}^3+3\mathrm{x}+{4\mathrm{y}}^3$$$$2^{\mathrm{nd}}\mathrm{order}\mathrm{f}\left({\mathrm{y}}^{{}^{″}}\right)=\frac{{\mathrm{d}}^2\mathrm{f}}{{\mathrm{dy}}^2}={12\mathrm{y}}^2.$$

Commented by Rio Michael last updated on 11/Apr/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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