Question

# Find all second-order partial derivatives for the following. k(x,y)=\frac{-7x}{2x+3y}

Derivatives
Find all second-order partial derivatives for the following.
$$\displaystyle{k}{\left({x},{y}\right)}={\frac{{-{7}{x}}}{{{2}{x}+{3}{y}}}}$$

2021-05-26
Step 1
Find all second-order partial derivatives for the following.
$$\displaystyle{k}{\left({x},{y}\right)}={\frac{{-{7}{x}}}{{{2}{x}+{3}{y}}}}$$
We first find the derivative with respect to x first,
$$\displaystyle{k}_{{{x}}}{\left({x},{y}\right)}={\frac{{{\left({2}{x}+{3}{y}\right)}{\left(-{7}\right)}-{\left(-{7}{x}\right)}{\left({2}\right)}}}{{{\left({2}{x}+{3}{y}\right)}^{{{2}}}}}}$$
$$\displaystyle{k}_{{{x}}}{\left({x},{y}\right)}={\frac{{-{14}{x}-{21}{y}+{14}{x}}}{{{\left({2}{x}+{3}{y}\right)}^{{{2}}}}}}$$
$$\displaystyle{k}_{{{x}}}{\left({x},{y}\right)}={\frac{{-{21}{y}}}{{{\left({2}{x}+{3}{y}\right)}^{{{2}}}}}}$$
$$\displaystyle{k}_{{\times}}{\left({x},{y}\right)}={\frac{{{\left({2}{x}+{3}{y}\right)}^{{{2}}}{\left({0}\right)}-{\left(-{21}{y}\right)}{2}{\left({2}{x}+{3}{y}\right)}{2}}}{{{\left({2}{x}+{3}{y}\right)}^{{{4}}}}}}$$
$$\displaystyle{k}_{{\times}}{\left({x},{y}\right)}={\frac{{{84}{y}}}{{{\left({2}{x}+{3}{y}\right)}^{{{3}}}}}}$$
Step 2
We find the derivative with respect to y,
$$\displaystyle{k}_{{{y}}}{\left({x},{y}\right)}={\frac{{{\left({2}{x}+{3}{y}\right)}{\left({0}\right)}-{\left(-{7}{x}\right)}{\left({3}\right)}}}{{{\left({2}{x}+{3}{y}\right)}^{{{2}}}}}}$$
$$\displaystyle{k}_{{{y}}}{\left({x},{y}\right)}={\frac{{{21}{x}}}{{{\left({2}{x}+{3}{y}\right)}^{{{2}}}}}}$$
$$\displaystyle{k}_{{{y}{y}}}{\left({x},{y}\right)}={\frac{{{\left({2}{x}+{3}{y}\right)}^{{{2}}}{\left({0}\right)}-{\left({21}{x}\right)}{2}{\left({2}{x}+{3}{y}\right)}{3}}}{{{\left({2}{x}+{3}{y}\right)}^{{{4}}}}}}$$
$$\displaystyle{k}_{{{y}{y}}}{\left({x},{y}\right)}={\frac{{-{126}{x}}}{{{\left({2}{x}+{3}{y}\right)}^{{{3}}}}}}$$