facas9

2021-05-17

Find the first partial derivatives.

$g(x,y)={e}^{\frac{x}{y}}$

Clelioo

Skilled2021-05-18Added 88 answers

Calculation:

Consider the provided function is,

$g(x,y)={e}^{\frac{x}{y}}$

Partially derivative of the function$g(x,y)={e}^{\frac{x}{y}}$ with respect to x.

${g}_{x}(x,y)=\frac{\partial}{\partial x}\left({e}^{\frac{x}{y}}\right)$

$={e}^{\frac{x}{y}}\left(\frac{1}{y}\right)$

$=\frac{1}{y}{e}^{\frac{x}{y}}$

Partially derivative of the function$g(x,y)={e}^{\frac{x}{y}}$ with respect to x.

${g}_{y}(x,y)=\frac{\partial}{\partial y}\left({e}^{\frac{x}{y}}\right)$

$={e}^{\frac{x}{y}}\left(\frac{x}{{y}^{2}}\right)$

$=\frac{-x}{{y}^{2}}{e}^{\frac{x}{y}}$

Therefore, the first partial derivatives for the function$g(x,y)={e}^{\frac{x}{y}}\text{}are\text{}{g}_{x}(x,y)=\frac{1}{y}{e}^{\frac{x}{y}}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{g}_{y}(x,y)=-\frac{x}{{y}^{2}}{e}^{\frac{x}{y}}$

Consider the provided function is,

Partially derivative of the function

Partially derivative of the function

Therefore, the first partial derivatives for the function

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