Find the derivative of \( y \) with re \[ y=\ln \left(\tan ^{-1}\left(4 x^{3}\right)\right) \] \( \frac{d y}{d x}=\square \)

To find the derivative of \( y = \ln\left(\tan^{-1}(4x^3)\right) \), we will use the chain rule and the derivative of the inverse tangent function. First, we identify the outer function and the inner function: - Outer function: \( u = \ln(v) \) where \( v = \tan^{-1}(4x^3) \) - Inner function: \( v = \tan^{-1}(u) \) Now, we differentiate: 1. The derivative of \( \ln(v) \) is \( \frac{1}{v} \frac{dv}{dx} \). 2. The derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \), where \( u = 4x^3 \). Next, we calculate \( \frac{du}{dx} \): \[ \frac{du}{dx} = \frac{d}{dx}(4x^3) = 12x^2 \] Now we calculate \( \frac{dv}{dx} \): \[ \frac{dv}{dx} = \frac{1}{1 + (4x^3)^2} \cdot 12x^2 = \frac{12x^2}{1 + 16x^6} \] Putting it all together, we have: \[ \frac{dy}{dx} = \frac{1}{v} \cdot \frac{dv}{dx} = \frac{1}{\tan^{-1}(4x^3)} \cdot \frac{12x^2}{1 + 16x^6} \] Thus, the final expression for the derivative is: \[ \frac{dy}{dx} = \frac{12x^2}{(1 + 16x^6) \tan^{-1}(4x^3)} \]