Find both first partial derivatives. h(x, y) = e^(-(x2+y2))

Find both first partial derivatives. h(x, y) = e^(-(x2+y2))

Question
Derivatives
asked 2020-10-27
Find both first partial derivatives. \(\displaystyle{h}{\left({x},{y}\right)}={e}^{{-{\left({x}{2}+{y}{2}\right)}}}\)

Answers (1)

2020-10-28
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,
\(\displaystyle\frac{{\partial{h}}}{{\partial{x}}}={e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\times{\left(-{2}{y}\right)}\)
\(\displaystyle=-{2}{y}{e}^{{-{\left({x}^{{2}}+{y}^{{2}}\right)}}}\)
0

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