Find the indicated derivative. \[ \frac{d y}{d x} \text { for } y=3 e^{x}\left(x^{3}+1\right) \] \( \frac{d y}{d x}=\square \)

We start with the function \[ y = 3e^x\left(x^3+1\right). \] We need to compute \(\frac{dy}{dx}\). Notice that \(y\) is the product of two functions: \[ u(x)=3e^x \quad \text{and} \quad v(x)=x^3+1. \] Using the product rule, which states that \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x)+u(x)v'(x), \] we first compute the derivatives of \(u(x)\) and \(v(x)\). 1. Differentiate \(u(x)\): \[ u(x)=3e^x \quad \Rightarrow \quad u'(x)=3e^x. \] 2. Differentiate \(v(x)\): \[ v(x)=x^3+1 \quad \Rightarrow \quad v'(x)=3x^2. \] Now, applying the product rule: \[ \frac{dy}{dx}= u'(x)v(x)+u(x)v'(x)= 3e^x\left(x^3+1\right) + 3e^x\left(3x^2\right). \] Factor the common term \(3e^x\): \[ \frac{dy}{dx}= 3e^x\left[ \left(x^3+1\right)+ 3x^2\right]. \] Simplify the expression inside the bracket: \[ \left(x^3+1\right)+ 3x^2 = x^3+3x^2+1. \] Thus, the derivative is: \[ \boxed{\frac{dy}{dx}=3e^x\left(x^3+3x^2+1\right)}. \]