Find the derivative of the function. \( y=3 \tan ^{-1}\left(x+\sqrt{1+x^{2}}\right) \) \( y^{\prime}=\square \)

Let \[ y=3\,\tan^{-1}\Bigl(x+\sqrt{1+x^2}\Bigr). \] We want to find \(y'\). One way to do this is to differentiate directly; an alternative is to use a substitution that simplifies the expression. We will use a hyperbolic substitution. **Step 1. Use a Hyperbolic Substitution** Let \[ x=\sinh t. \] Then \[ \sqrt{1+x^2}=\sqrt{1+\sinh^2t}=\cosh t. \] Thus, \[ x+\sqrt{1+x^2} = \sinh t+\cosh t = e^t. \] So the function becomes \[ y=3\,\tan^{-1}\Bigl(e^t\Bigr). \] **Step 2. Differentiate with Respect to \(t\)** Differentiate \(y\) with respect to \(t\): \[ \frac{dy}{dt}=3\cdot\frac{d}{dt}\Bigl[\tan^{-1}\bigl(e^t\bigr)\Bigr]=3\cdot\frac{1}{1+e^{2t}}\cdot e^t. \] Thus, \[ \frac{dy}{dt}=\frac{3e^t}{1+e^{2t}}. \] **Step 3. Express \(\frac{dy}{dx}\) in Terms of \(t\)** We have \(x=\sinh t\), so \[ \frac{dx}{dt}=\cosh t. \] Then, by the chain rule, \[ \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}=\frac{3e^t}{1+e^{2t}}\cdot\frac{1}{\cosh t}. \] Notice that \[ \frac{e^t}{1+e^{2t}} = \frac{1}{e^{-t}+e^t}=\frac{1}{2\cosh t}, \] since \[ e^{-t}+e^{t}=2\cosh t. \] Thus, \[ \frac{dy}{dx}=\frac{3}{2\cosh t}\cdot\frac{1}{\cosh t} = \frac{3}{2\cosh^2t}. \] **Step 4. Rewrite in Terms of \(x\)** Recall that \[ \cosh t=\sqrt{1+\sinh^2t}=\sqrt{1+x^2}. \] So, \[ \cosh^2t=1+x^2. \] Therefore, \[ y'=\frac{3}{2(1+x^2)}. \] **Final Answer** \[ y'=\frac{3}{2(1+x^2)}. \]