Mylo O'Moore

2021-10-08

The following questions are about the function $f(x,y)={x}^{2}{e}^{2xy}$ .

Find the partial derivatives,${f}_{x}(x,y)\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{f}_{y}(x,y)$ .

Find the partial derivatives,

Arnold Odonnell

Skilled2021-10-09Added 109 answers

Step 1

If f(x,y) is a function of two variables x and y, then

$f}_{x}(x,y)=\frac{\partial f}{\partial x$

$f}_{y}(x,y)=\frac{\partial f}{\partial y$

The equation of the tangent plane of function f at the point$({x}_{0},{y}_{0})$ is

$z=f({x}_{0},{y}_{0})+{f}_{x}({x}_{0},{y}_{0})(x-{x}_{0})+{f}_{y}({x}_{0},{y}_{0})(y-{y}_{0})$

Step 2

The given function is

$f(x,y)={x}^{2}{e}^{2xy}$

Differentiate the given function partially with respect to x.

$f}_{x}(x,y)={x}^{2}\frac{\partial}{\partial x}\left({e}^{2xy}\right)+{e}^{2xy}\frac{\partial}{\partial x}{x}^{2$

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

$=2x{e}^{2xy}(xy+1)$

Differentiate the given function partially with respect to y.

${f}_{y}(x,y)={x}^{2}\frac{\partial}{\partial y}\left({e}^{2xy}\right)$

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

$=2{x}^{3}{e}^{2xy}$

If f(x,y) is a function of two variables x and y, then

The equation of the tangent plane of function f at the point

Step 2

The given function is

Differentiate the given function partially with respect to x.

Differentiate the given function partially with respect to y.

Jeffrey Jordon

Expert2022-08-15Added 2607 answers

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