 # Find the second partial derivatives. f(x,y)=x^{4}y-3x^{5}y^{2}f_{xx}(x,y)=f_{xy}(x,y)=f_{yx}(x,y)=f_{yy}(x,y) Tabansi 2021-11-08 Answered

Find all the second partial derivatives.
$f\left(x,y\right)={x}^{4}y-3{x}^{5}{y}^{2}$
${f}_{xx}\left(x,y\right)=$
${f}_{xy}\left(x,y\right)=$
${f}_{yx}\left(x,y\right)=$
${f}_{yy}\left(x,y\right)=$

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Step 1
Given:
$f\left(x,y\right)={x}^{4}y-3{x}^{5}{y}^{2}$
Step 2
$f\left(x,y\right)={x}^{4}y-3{x}^{5}{y}^{2}$
${f}_{x}\left(x,y\right)=\frac{\partial }{\partial x}\left(f\left(x,y\right)\right)=4{x}^{3}-15{x}^{4}{y}^{2}$
${f}_{xy}\left(x,y\right)=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)$
$=\frac{\partial }{\partial y}\left(4{x}^{3}-15{x}^{4}{y}^{2}\right)$
$=\frac{\partial }{\partial y}\left(4{x}^{3}\right)-\frac{\partial }{\partial y}\left(15{x}^{4}{y}^{2}\right)$
$=0-15{x}^{4}\left(2y\right)$
${f}_{xy}\left(x,y\right)=-30{x}^{4}y$
${f}_{xx}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)$
$=\frac{\partial }{\partial x}\left(4{x}^{3}-15{x}^{4}{y}^{2}\right)$
$=\frac{\partial }{\partial x}\left(4{x}^{3}\right)-\frac{\partial }{\partial x}\left(15{x}^{4}{y}^{2}\right)$
${f}_{xx}=12{x}^{2}-60{x}^{3}{y}^{2}$
${f}_{y}=\frac{\partial }{\partial y}\left({x}^{4}y-3{x}^{5}{y}^{2}\right)$
$=\frac{\partial }{\partial y}\left({x}^{4}y\right)-\frac{\partial }{\partial y}\left(3{x}^{5}{y}^{2}\right)$
$={x}^{4}-3{x}^{5}\left(2y\right)$
${f}_{y}={x}^{4}-6{x}^{5}y$
${f}_{yx}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)$

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