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

To find the derivative of the function \( f(x)=\left(25+x^{2}\right) \sqrt{\arctan \left(\frac{x}{5}\right)} \), we will use the product rule, which states that if you have two functions multiplied together, the derivative is given by \( (uv)' = u'v + uv' \). Let: - \( u = 25 + x^2 \) - \( v = \sqrt{\arctan\left(\frac{x}{5}\right)} \) First, we compute the derivatives \( u' \) and \( v' \): 1. \( u' = 2x \) 2. For \( v' \), we apply the chain rule: - \( v = (\arctan(\frac{x}{5}))^{1/2} \) - \( v' = \frac{1}{2} (\arctan(\frac{x}{5}))^{-1/2} \cdot \frac{d}{dx}(\arctan(\frac{x}{5})) \) - From the derivative of arctangent, we have \( \frac{d}{dx}\left(\arctan\left(\frac{x}{5}\right)\right) = \frac{1}{1 + \left(\frac{x}{5}\right)^2} \cdot \frac{1}{5} = \frac{5}{25 + x^2} \) Thus, \( v' = \frac{1}{2} (\arctan(\frac{x}{5}))^{-1/2} \cdot \frac{5}{25 + x^2} \). Now applying the product rule: \[ f'(x) = u'v + uv' = (2x) \sqrt{\arctan\left(\frac{x}{5}\right)} + (25 + x^2) \left( \frac{1}{2} \left(\arctan\left(\frac{x}{5}\right)\right)^{-1/2} \cdot \frac{5}{25 + x^2} \right) \] This simplifies to: \[ f'(x) = 2x \sqrt{\arctan\left(\frac{x}{5}\right)} + \frac{(25 + x^2) \cdot 5}{2(25 + x^2) \sqrt{\arctan\left(\frac{x}{5}\right)}} \] Finally, you can factor and simplify as necessary. So, the derivative \( f^{\prime}(x) \) is: \[ f^{\prime}(x) = 2x \sqrt{\arctan\left(\frac{x}{5}\right)} + \frac{5}{2} \sqrt{\arctan\left(\frac{x}{5}\right)} \] You can combine these terms if needed, but the important part is that you've successfully found the derivative!