Mylo O'Moore

2021-03-09

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

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

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