Determine $$ f_{x} $$ und $$ f y $$ of the given function $$ f(x, y)=\left(x^{5} \ln 2 x\right) y^{8} $$

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To find the partial derivatives $$ f_x $$ and $$ f_y $$ of the function
$$
f(x, y) = \left(x^{5} \ln(2x)\right) y^{8},
$$
we will differentiate with respect to $$ x $$ and $$ y $$ separately, treating the other variable as a constant in each case.

### Partial Derivative with Respect to $$ x $$ ($$ f_x $$)

$$
f(x, y) = x^{5} \ln(2x) \cdot y^{8}
$$

To differentiate $$ f $$ with respect to $$ x $$, treat $$ y^8 $$ as a constant:

$$
f_x = y^{8} \cdot \frac{d}{dx}\left(x^{5} \ln(2x)\right)
$$

Use the **product rule** for differentiation, where if $$ u = x^5 $$ and $$ v = \ln(2x) $$, then:

$$
\frac{d}{dx}(u \cdot v) = u' \cdot v + u \cdot v'
$$

Compute $$ u' $$ and $$ v' $$:

$$
u' = \frac{d}{dx}(x^5) = 5x^4
$$
$$
v' = \frac{d}{dx}(\ln(2x)) = \frac{1}{2x} \cdot 2 = \frac{1}{x}
$$

Applying the product rule:

$$
\frac{d}{dx}\left(x^{5} \ln(2x)\right) = 5x^{4} \ln(2x) + x^{5} \cdot \frac{1}{x} = 5x^{4} \ln(2x) + x^{4}
$$

Factor out $$ x^4 $$:

$$
f_x = y^{8} \left(5x^{4} \ln(2x) + x^{4}\right) = y^{8} x^{4} \left(5 \ln(2x) + 1\right)
$$

### Partial Derivative with Respect to $$ y $$ ($$ f_y $$)

$$
f(x, y) = x^{5} \ln(2x) \cdot y^{8}
$$

To differentiate $$ f $$ with respect to $$ y $$, treat $$ x^5 \ln(2x) $$ as a constant:

$$
f_y = x^{5} \ln(2x) \cdot \frac{d}{dx}(y^{8}) = x^{5} \ln(2x) \cdot 8y^{7} = 8x^{5} \ln(2x) y^{7}
$$

### Summary

The partial derivatives of $$ f(x, y) = \left(x^{5} \ln(2x)\right) y^{8} $$ are:

$$
f_x = y^{8} x^{4} \left(5 \ln(2x) + 1\right)
$$
$$
f_y = 8x^{5} \ln(2x) y^{7}
$$
The partial derivatives are: $$ f_x = y^{8} x^{4} \left(5 \ln(2x) + 1\right) $$ $$ f_y = 8x^{5} \ln(2x) y^{7} $$

Determine und of the
given function

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