# Find the four second partial derivatives of f(x, y) = x^2y^3.

Find the four second partial derivatives of $f\left(x,y\right)={x}^{2}{y}^{3}$.
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Step 1
Given function:
$f\left(x,y\right)={x}^{2}{y}^{3}$
The four-second partial derivatives of $f\left(x,y\right):{f}_{xx},{f}_{yy},{f}_{xy},{f}_{yx}$
Step 2
Now,
Apply the Power Rule: $\frac{d}{dx}\left({x}^{a}\right)=a\cdot {x}^{a-1}$ to find the partial derivatives:
${f}_{xx}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial x}\right)$
$=\frac{\partial }{\partial x}\left(\frac{\partial {x}^{2}{y}^{3}}{\partial x}\right)$
$=\left(\partial \frac{2x{y}^{3}}{\partial x}=2{y}^{3}$
${f}_{yy}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial y}\right)$
$=\frac{\partial }{\partial y}\left(\frac{\partial {x}^{2}{y}^{3}}{\partial }y\right)\right)$
$=\left(\partial \frac{3{x}^{2}{y}^{2}}{\partial }y\right)$
$=3{x}^{2}×2y$
$=6{x}^{2}y$
${f}_{xy}=\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)$
$=\frac{\partial }{\partial y}\frac{\partial {x}^{2}{y}^{3}}{\partial x}\right)$
$=\left(\partial \frac{2x{y}^{3}}{\partial y}\right)$
$=6{x}^{2}$
${f}_{yx}=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right)$
$=\frac{\partial }{\partial x}\left(\frac{\partial {x}^{2}{y}^{3}}{\partial y}\right)$
$=\left(\partial \frac{3{x}^{2}{y}^{2}}{\partial x}\right)$
$=6x{y}^{2}$