Let $$ f(x)=2 x^{3} \ln (x)+\frac{5}{6} x^{6} $$ $$ f^{\prime}(x)=\square $$

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We start with
$$
f(x)=2x^3\ln(x)+\frac{5}{6}x^6.
$$

### Step 1. Differentiate the first term $$2x^3\ln(x)$$
Use the product rule for the term $$x^3\ln(x)$$. If 
$$
u(x)=x^3 \quad \text{and} \quad v(x)=\ln(x),
$$
then
$$
u'(x)=3x^2 \quad \text{and} \quad v'(x)=\frac{1}{x}.
$$
The derivative of $$x^3\ln(x)$$ is
$$
\frac{d}{dx}\left(x^3\ln(x)\right)=u'(x)v(x)+u(x)v'(x)=3x^2\ln(x)+x^3\left(\frac{1}{x}\right)=3x^2\ln(x)+x^2.
$$
Multiplying by 2 gives:
$$
\frac{d}{dx}\left(2x^3\ln(x)\right)=2\left[3x^2\ln(x)+x^2\right]=6x^2\ln(x)+2x^2.
$$

### Step 2. Differentiate the second term $$\frac{5}{6}x^6$$
Use the power rule:
$$
\frac{d}{dx}\left(x^6\right)=6x^5.
$$
Thus,
$$
\frac{d}{dx}\left(\frac{5}{6}x^6\right)=\frac{5}{6}\cdot6x^5=5x^5.
$$

### Step 3. Combine the derivatives
Adding the derivatives from Steps 1 and 2, we obtain:
$$
f'(x)=6x^2\ln(x)+2x^2+5x^5.
$$

Thus, the derivative is:
$$
f'(x)=6x^2\ln(x)+2x^2+5x^5.
$$
$$ f'(x) = 6x^2 \ln(x) + 2x^2 + 5x^5 $$

Let \( f(x)=2 x^{3} \ln (x)+\frac{5}{6} x^{6} \) \( f^{\prime}(x)=\square \)

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