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

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