Find the derivative of the function below. \[ f(x)=\left(25+x^{2}\right) \sqrt{\arctan \left(\frac{x}{5}\right)} \] \( f^{\prime}(x)=\square \)

We write \[ f(x)=\bigl(25+x^2\bigr)\sqrt{\arctan\left(\frac{x}{5}\right)} \] as a product of two functions: \[ u(x)=25+x^2 \quad \text{and} \quad v(x)=\sqrt{\arctan\left(\frac{x}{5}\right)}=\left(\arctan\left(\frac{x}{5}\right)\right)^{\frac{1}{2}}. \] Then, by the product rule, \[ f'(x)=u'(x)v(x)+u(x)v'(x). \] **Step 1. Differentiating \(u(x)\):** \[ u(x)=25+x^2 \quad \Longrightarrow \quad u'(x)=2x. \] **Step 2. Differentiating \(v(x)\):** Express \[ v(x)=\left(\arctan\left(\frac{x}{5}\right)\right)^{\frac{1}{2}}. \] Using the chain rule with the outer function \(g(z)=z^{\frac{1}{2}}\) and the inner function \(h(x)=\arctan\left(\frac{x}{5}\right)\), we obtain \[ v'(x)=\frac{1}{2}\left(\arctan\left(\frac{x}{5}\right)\right)^{-\frac{1}{2}} \cdot h'(x). \] Now, differentiate the inner function \(h(x)=\arctan\left(\frac{x}{5}\right)\). Recall that the derivative of \(\arctan(u)\) is \[ \frac{d}{dx}\arctan(u)=\frac{u'}{1+u^2}. \] Here, \(u(x)=\frac{x}{5}\) so that \(u'(x)=\frac{1}{5}\). Therefore, \[ h'(x)=\frac{\frac{1}{5}}{1+\left(\frac{x}{5}\right)^2}=\frac{\frac{1}{5}}{1+\frac{x^2}{25}}. \] Simplify the denominator: \[ 1+\frac{x^2}{25}=\frac{25+x^2}{25}, \] so \[ h'(x)=\frac{1/5}{(25+x^2)/25}=\frac{1}{5}\cdot\frac{25}{25+x^2}=\frac{5}{25+x^2}. \] Thus, \[ v'(x)=\frac{1}{2}\left(\arctan\left(\frac{x}{5}\right)\right)^{-\frac{1}{2}} \cdot \frac{5}{25+x^2}=\frac{5}{2(25+x^2)\sqrt{\arctan\left(\frac{x}{5}\right)}}. \] **Step 3. Combining the results:** Substitute back into the product rule: \[ f'(x)=2x\sqrt{\arctan\left(\frac{x}{5}\right)}+\left(25+x^2\right)\cdot\frac{5}{2(25+x^2)\sqrt{\arctan\left(\frac{x}{5}\right)}}. \] Notice that the factor \(25+x^2\) cancels in the second term: \[ \left(25+x^2\right)\cdot\frac{5}{2(25+x^2)\sqrt{\arctan\left(\frac{x}{5}\right)}}=\frac{5}{2\sqrt{\arctan\left(\frac{x}{5}\right)}}. \] Thus, the derivative is \[ f'(x)=2x\sqrt{\arctan\left(\frac{x}{5}\right)}+\frac{5}{2\sqrt{\arctan\left(\frac{x}{5}\right)}}. \] So, the final answer is: \[ \boxed{ f'(x)=2x\sqrt{\arctan\left(\frac{x}{5}\right)}+\frac{5}{2\sqrt{\arctan\left(\frac{x}{5}\right)}} }. \]