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

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

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

Commented by niroj last updated on 10/Apr/20

    f(x,y)=  x^3 y+3xy+y^4        D.w.r.to x    1^(st) order f(x^′ )=(df/dx)= 3x^2 y+3y   2^(nd) order f(x^(′′) )= (d^2 f/dx^2 )= 6xy.    Again,   f(y)=x^3 y+3xy+y^4      D.w.r.to y    1^(st) order f(y^′ )= (df/dy)= x^3 +3x+4y^3    2^(nd)  order f(y^(′′) )= (d^2 f/dy^2 )= 12y^2 .

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

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

thank you sir

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

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