Question

# Given f(x,y)=2x^{2}-xy^{3}+4y^{6}, findf_{xx}(x,y)=f_{xy}(x,y)=

Derivatives

Given $$\displaystyle{f{{\left({x},{y}\right)}}}={2}{x}^{{{2}}}-{x}{y}^{{{3}}}+{4}{y}^{{{6}}}$$, find
$$\displaystyle{{f}_{{xx}}{\left({x},{y}\right)}}=$$
$$\displaystyle{{f}_{{{x}{y}}}{\left({x},{y}\right)}}=$$

2021-06-08

Step 1
Consider the given function.
$$\displaystyle{f{{\left({x},{y}\right)}}}={2}{x}^{{{2}}}-{x}{y}^{{{3}}}+{4}{y}^{{{6}}}$$
The objective of the question is to find the value of $$\displaystyle{{f}_{{xx}}{\left({x},{y}\right)}}\ {\quad\text{and}\quad}\ {{f}_{{{x}{y}}}{\left({x},{y}\right)}}$$.
These are the second order partial derivatives.
Step 2
Partially differentiate the function with respect to x.
$$\displaystyle{{f}_{{{x}}}{\left({x},{y}\right)}}={2}\cdot{3}{x}^{{{3}-{1}}}-{1}\cdot{y}^{{{3}}}+{0}$$
$$\displaystyle={6}{x}^{{{2}}}-{y}^{{{3}}}$$
Partially differentiate again with respect to x to find $$\displaystyle{{f}_{{xx}}{\left({x},{y}\right)}}$$.
$$\displaystyle{{f}_{{xx}}{\left({x},{y}\right)}}={6}\cdot{2}{x}-{0}={12}{x}$$
Partially differentiate the function with respect to y.
$$\displaystyle{{f}_{{{y}}}{\left({x},{y}\right)}}={0}-{x}\cdot{3}{y}^{{{3}-{1}}}+{4}\cdot{6}{y}^{{{6}-{1}}}$$
$$\displaystyle=-{3}{x}{y}^{{{2}}}+{24}{y}^{{{5}}}$$
Partially differentiate again with respect to x to find $$\displaystyle{{f}_{{{x}{y}}}{\left({x},{y}\right)}}$$.
$$\displaystyle{{f}_{{{x}{y}}}{\left({x},{y}\right)}}=-{3}{y}^{{{2}}}\cdot{1}+{0}$$
$$\displaystyle=-{3}{y}^{{{2}}}$$
Thus, the required partial derivatives are obtained.
$$\displaystyle{{f}_{{xx}}{\left({x},{y}\right)}}={12}{x}$$
$$\displaystyle{{f}_{{{x}{y}}}{\left({x},{y}\right)}}=-{3}{y}^{{{2}}}$$