# Find the first partial derivatives of the following functions. f(x,y)=xe^y

Question
Derivatives
Find the first partial derivatives of the following functions. $$\displaystyle{f{{\left({x},{y}\right)}}}={x}{e}^{{y}}$$

2021-03-10
Step 1
To find the first-order partial derivatives of the function,
$$\displaystyle{f{{\left({x},{y}\right)}}}={x}{e}^{{y}}$$
Step 2
The partial derivatives with respect to x is given by,
$$\displaystyle{{f}_{{x}}{\left({x},{y}\right)}}=\lim_{{{h}\rightarrow{0}}}\frac{{{f{{\left({x}+{h},{y}\right)}}}-{f{{\left({x},{y}\right)}}}}}{{h}}=\lim_{{{h}\rightarrow{0}}}\frac{{{\left({x}+{h}\right)}{e}^{{y}}-{x}{e}^{{y}}}}{{h}}=\lim_{{{h}\rightarrow{0}}}\frac{{{x}{e}^{{y}}+{h}{e}^{{y}}-{x}{e}^{{y}}}}{{h}}=\lim_{{{h}\rightarrow{0}}}\frac{{{h}{e}^{{y}}}}{{h}}$$
$$\displaystyle=\lim_{{{h}\rightarrow{0}}}{e}^{{y}}={e}^{{y}}$$
The partial derivatives with respect to y is given by,
$$\displaystyle{{f}_{{y}}{\left({x},{y}\right)}}=\lim_{{{k}\rightarrow{0}}}\frac{{{f{{\left({x},{y}+{k}\right)}}}-{f{{\left({x},{y}\right)}}}}}{{h}}=\lim_{{{k}\rightarrow{0}}}\frac{{{x}{e}^{{{y}+{k}}}-{x}{e}^{{y}}}}{{k}}=\lim_{{{k}\rightarrow{0}}}\frac{{{x}{e}^{{y}}{\left({e}^{{k}}-{1}\right)}}}{{k}}={x}{e}^{{y}}\lim_{{{k}\rightarrow{0}}}\frac{{{e}^{{k}}-{1}}}{{k}}$$
$$\displaystyle={x}{e}^{{y}}\lim_{{{k}\rightarrow{0}}}{e}^{{k}}={x}{e}^{{y}}{\left({1}\right)}={x}{e}^{{y}}$$

### Relevant Questions

Find the first partial derivatives of the following functions.
$$\displaystyle{f{{\left({x},{y}\right)}}}={\frac{{{x}^{{{2}}}}}{{{x}^{{{2}}}+{y}^{{{2}}}}}}$$
Find the first partial derivatives of the following functions.
$$\displaystyle{f{{\left({x},{y}\right)}}}={4}{x}^{{{3}}}{y}^{{{2}}}+{3}{x}^{{{2}}}{y}^{{{3}}}+{10}$$
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$$\displaystyle{f{{\left({x},{y}\right)}}}={\left({3}{x}{y}+{4}{y}^{{{2}}}+{1}\right)}^{{{5}}}$$
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Find the first partial derivatives of the given functions with respect to each independent variable.
$$\displaystyle{f{{\left({x},{y}\right)}}}={\left({2}{x}-{4}\right)}^{{{4}}}$$
Find the first partial derivatives of the following functions.
$$\displaystyle{h}{\left({x},{y}\right)}={x}-\sqrt{{{x}^{{{2}}}-{4}{y}}}$$
$$\displaystyle{g{{\left({x},{y}\right)}}}={y}{{\sin}^{{-{1}}}\sqrt{{{x}{y}}}}$$
$$\displaystyle{f{{\left({x},{y}\right)}}}={e}^{{{2}{x}{y}}}$$
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$$\displaystyle{f{{\left({x},{y}\right)}}}={y}{\cos{{\left({x}^{{{2}}}+{y}^{{{2}}}\right)}}}$$