Question

find the partial derivatives of the function

$f(x,y)=xye^{-3y}$

$f_x(x,y)=ye^{-3y}$

$f_y(x,y)=-3xe^{-3y}$

$f_{xy}(x,y)=-3e^{-3y}$

$f_{yx}(x,y)=-3e^{-3y}$

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Explanation:

Step1: Find $f_x(x,y)$

Treat $y$ as a constant. Derivative of $xye^{-3y}$ with respect to $x$: $\frac{\partial}{\partial x}(xye^{-3y})=ye^{-3y}$.

Step2: Find $f_y(x,y)$

Treat $x$ as a constant. Use product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = xy$ and $v = e^{-3y}$. $u^\prime=x$, $v^\prime=- 3e^{-3y}$. Then $f_y(x,y)=x(e^{-3y}-3ye^{-3y})=xe^{-3y}(1 - 3y)$. The given $f_y(x,y)=-3xe^{-3y}$ is incorrect.

Step3: Find $f_{xy}(x,y)$

Differentiate $f_x(x,y)=ye^{-3y}$ with respect to $y$. Using product - rule with $u = y$ and $v = e^{-3y}$, $u^\prime = 1$, $v^\prime=-3e^{-3y}$, we get $f_{xy}(x,y)=e^{-3y}-3ye^{-3y}$. The given $f_{xy}(x,y)=-3e^{-3y}$ is incorrect.

Step4: Find $f_{yx}(x,y)$

Differentiate $f_y(x,y)=xe^{-3y}(1 - 3y)$ with respect to $x$. Treat $y$ as a constant, we get $f_{yx}(x,y)=e^{-3y}(1 - 3y)$. The given $f_{yx}(x,y)=-3e^{-3y}$ is incorrect.

Answer:

$f_x(x,y)=ye^{-3y}$
$f_y(x,y)=xe^{-3y}(1 - 3y)$
$f_{xy}(x,y)=e^{-3y}-3ye^{-3y}$
$f_{yx}(x,y)=e^{-3y}(1 - 3y)$