2021-03-02

COnsider the multivariable function $g\left(x,y\right)={x}^{2}-3{y}^{4}{x}^{2}+\mathrm{sin}\left(xy\right)$. Find the following partial derivatives: ${g}_{x}.{g}_{y},{g}_{xy},g\left(×\right),g\left(yy\right)$.

### Answer & Explanation

Theodore Schwartz

${g}_{x}=\frac{\partial }{\partial x}\left({x}^{2}-3{y}^{4}{x}^{2}+\mathrm{sin}x\right)\right)$ [treat y as cons $\mathrm{tan}t$]
$=2x-6x{y}^{4}+y\mathrm{cos}xy$
Again
${g}_{y}=\frac{\partial }{\partial y}\left({x}^{2}-3{y}^{4}{x}^{2}+\mathrm{sin}xy\right)$ [treat x as cons $\mathrm{tan}t$]
$=-12{x}^{2}{y}^{3}+x\mathrm{cos}xy$
Now
${g}_{xx}=\frac{\partial }{\partial x}\left({g}_{x}\right)=\frac{\partial }{\partial x}\left(2x-6x{y}^{4}+y\mathrm{cos}xy\right)$
$=2-6{y}^{4}-{y}^{2}\mathrm{sin}xy$
Again
${g}_{yy}=\frac{\partial }{\partial y}\left(-12{x}^{2}{y}^{3}+x\mathrm{cos}xy\right)$
$=-36{x}^{2}{y}^{2}-{x}^{2}\mathrm{sin}xy$
Also
${g}_{xy}=\frac{\partial }{\partial y}\left({g}_{x}\right)$
$=\frac{\partial }{\partial y}\left(2x-6x{y}^{4}+y\mathrm{cos}xy\right)$
$=-24x{y}^{3}-{y}^{2}\mathrm{sin}xy+\mathrm{cos}xy$

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