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

Ask by Conner John. in the United States
Mar 16,2025

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The derivative of the function \( f(x) = (25 + x^2) \sqrt{\arctan\left(\frac{x}{5}\right)} \) is: \[ f'(x) = 2x \sqrt{\arctan\left(\frac{x}{5}\right)} + \frac{5}{2 \sqrt{\arctan\left(\frac{x}{5}\right)}} \]

Solución

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: \[ \boxed{ f'(x)=2x\sqrt{\arctan\left(\frac{x}{5}\right)}+\frac{5}{2\sqrt{\arctan\left(\frac{x}{5}\right)}} }. \]

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Mind Expander

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!

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