Nannie Mack

2021-03-02

Find all first and the second partial derivatives.
$f\left(x,y\right)=2{x}^{5}{y}^{2}+{x}^{2}y$

Leonard Stokes

Step 1 Let us find first and second order partial derivatives.
First order partial derivatives:
${f}_{x}=\frac{\partial f}{\partial x}{f}_{y}=\frac{\partial f}{\partial y}$
Second order partial derivatives:
${f}_{xx}=\frac{{\partial }^{2}f}{\partial {x}^{2}}=\frac{\partial }{\partial x}\left(\frac{\partial }{\partial x}\right),{f}_{yy}=\frac{{\partial }^{2}f}{\partial {y}^{2}}=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}\right)$
Step 2 Answer will be at end of step 2
$f\left(x,y\right)=2{x}^{5}{y}^{2}+{x}^{2}y$
First ordrer partial derivative:
${f}_{x}=\frac{\partial f}{\partial x}=2.\left(5\right){x}^{5-1}{y}^{2}+2{x}^{2-1}y$
${f}_{x}=10{x}^{4}{y}^{2}+2xy$
${f}_{y}=\frac{\partial f}{\partial y}=2{x}^{5}.\left(2\right){y}^{2-1}+{x}^{2}$
${f}_{y}=4{x}^{5}y+{x}^{2}$
Second order partial derivative:
${f}_{xx}\frac{\partial }{\partial x}\left(10{x}^{4}{y}^{2}+2xy\right)$
${f}_{xx}4.\left(10\right){x}^{3}{y}^{2}+2y$
${f}_{xx}=40{x}^{3}{y}^{2}+2y$
${f}_{yy}=\frac{\partial }{\partial y}\left(4{x}^{5}y+{x}^{2}\right)$
${f}_{yy}=4{x}^{5}$
Step 3
Result:
${f}_{x}=10{x}^{4}{y}^{2}+2xy$
${f}_{y}=4{x}^{5}y+{x}^{2}$
${f}_{xx}=40{x}^{3}{y}^{2}+2y$
${f}_{yy}=4{x}^{5}$

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