Step 1

The given function is \(\displaystyle{h}{\left({x},{y}\right)}={e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\).

To find the partial derivatives.

Solution:

The given function is \(\displaystyle{h}{\left({x},{y}\right)}={e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\).

Now differentiate partially above function with respect to x,

\(\displaystyle\frac{{\partial{h}}}{{\partial{x}}}={e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\times{\left(-{2}{x}\right)}\)

\(\displaystyle=-{2}{x}{e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\)

Step 2

Now differentiate partially above function with respect to y,

\(\displaystyle\frac{{\partial{h}}}{{\partial{x}}}={e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\times{\left(-{2}{y}\right)}\)

\(\displaystyle=-{2}{y}{e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\)

The given function is \(\displaystyle{h}{\left({x},{y}\right)}={e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\).

To find the partial derivatives.

Solution:

The given function is \(\displaystyle{h}{\left({x},{y}\right)}={e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\).

Now differentiate partially above function with respect to x,

\(\displaystyle\frac{{\partial{h}}}{{\partial{x}}}={e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\times{\left(-{2}{x}\right)}\)

\(\displaystyle=-{2}{x}{e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\)

Step 2

Now differentiate partially above function with respect to y,

\(\displaystyle\frac{{\partial{h}}}{{\partial{x}}}={e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\times{\left(-{2}{y}\right)}\)

\(\displaystyle=-{2}{y}{e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\)