𝑓(𝑥, 𝑦, 𝑢, 𝑣) = ln (𝑥𝑦) − 𝑣𝑒𝑢𝑦a.Find 𝑓𝑥 , 𝑓𝑦 , 𝑓𝑢 , 𝑎𝑛𝑑 𝑓𝑣

Partial Derivative with respect to $x$:

The partial derivative of $f$ with respect to $x$, denoted as $f_x$, is found by differentiating $f$ with respect to $x$ while treating $y$, $u$, and $v$ as constants.

$$f_x=\frac{\partial }{\partial x}(\ln (xy)-v{e}^{uy})$$

The derivative of $\ln (xy)$ with respect to $x$ is $\frac{1}{xy}\cdot y$ because the derivative of $\ln (x)$ is $\frac{1}{x}$, and by the chain rule, we multiply by the derivative of the inside function, which is $y$. The second term does not contain $x$, so its derivative with respect to $x$ is zero.

$$f_x=\frac{y}{xy}-0=\frac{1}{x}$$

Partial Derivative with respect to $y$:

The partial derivative of $f$ with respect to $y$, denoted as $f_y$, is found by differentiating $f$ with respect to $y$ while treating $x$, $u$, and $v$ as constants.

$$f_y=\frac{\partial }{\partial y}(\ln (xy)-v{e}^{uy})$$

The derivative of $\ln (xy)$ with respect to $y$ is $\frac{1}{xy}\cdot x$ for the same reason as above. For the second term, we use the product rule: the derivative of $-v{e}^{uy}$ with respect to $y$ is $-v\cdot u\cdot e^{uy}$, because the derivative of $e^{uy}$ with respect to $y$ is $u\cdot e^{uy}$ by the chain rule.

$$f_y=\frac{x}{xy}-vu\cdot e^{uy}=\frac{1}{y}-vu\cdot e^{uy}$$

Partial Derivative with respect to $u$:

The partial derivative of $f$ with respect to $u$, denoted as $f_u$, is found by differentiating $f$ with respect to $u$ while treating $x$, $y$, and $v$ as constants.

$$f_u=\frac{\partial }{\partial u}(\ln (xy)-v{e}^{uy})$$

The first term does not contain $u$, so its derivative with respect to $u$ is zero. For the second term, we again use the product rule: the derivative of $-v{e}^{uy}$ with respect to $u$ is $-vy\cdot e^{uy}$.

$$f_u=0-vy\cdot e^{uy}=-vy\cdot e^{uy}$$

Partial Derivative with respect to $v$:

The partial derivative of $f$ with respect to $v$, denoted as $f_v$, is found by differentiating $f$ with respect to $v$ while treating $x$, $y$, and $u$ as constants.

$$f_v=\frac{\partial }{\partial v}(\ln (xy)-v{e}^{uy})$$

The first term does not contain $v$, so its derivative with respect to $v$ is zero. The derivative of $-v{e}^{uy}$ with respect to $v$ is simply $-e^{uy}$ because $e^{uy}$ is treated as a constant with respect to $v$.

$$f_v=0-e^{uy}=-e^{uy}$$

In summary, the partial derivatives of the function $f(x,y,u,v)$ are:

$$f_x=\frac{1}{x},f_y=\frac{1}{y}-vu\cdot e^{uy},f_u=-vy\cdot e^{uy},f_v=-e^{uy}$$