Find the derivative of the function. \( y=3 \tan ^{-1}\left(x+\sqrt{1+x^{2}}\right) \) \( y^{\prime}=\square \)
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To find the derivative of the function \( y = 3 \tan^{-1}\left(x + \sqrt{1 + x^2}\right) \), we start by differentiating using the chain rule. First, we differentiate the outer function: \[ y' = 3 \cdot \frac{1}{1 + \left(x + \sqrt{1 + x^2}\right)^2} \cdot \frac{d}{dx}\left(x + \sqrt{1 + x^2}\right) \] Next, we differentiate the inner function: \[ \frac{d}{dx}\left(x + \sqrt{1 + x^2}\right) = 1 + \frac{x}{\sqrt{1 + x^2}} \] Now, let's combine these parts together: \[ y' = 3 \cdot \frac{1}{1 + \left(x + \sqrt{1 + x^2}\right)^2} \cdot \left(1 + \frac{x}{\sqrt{1 + x^2}}\right) \] To simplify, recall that \(1 + x + \sqrt{1 + x^2} = (1 + x)^2\). Thus: \[ y' = \frac{3\left(1 + \frac{x}{\sqrt{1 + x^2}}\right)}{1 + (x + \sqrt{1 + x^2})^2} \] And that is your derivative! Now, filling in \( y' =\) with your result gives you the complete answer.