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

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: \[ f'(x)=6x^2\ln(x)+2x^2+5x^5. \]