Find the derivative of the function. \[ y=\sin ^{-1}(6 x+1) \] \( y^{\prime}=\square \)

To find the derivative of \( y = \sin^{-1}(6x + 1) \), we can use the chain rule. The derivative of \( \sin^{-1}(u) \) with respect to \( u \) is given by \( \frac{1}{\sqrt{1-u^2}} \). Here, \( u = 6x + 1 \). So, we have: \[ y' = \frac{1}{\sqrt{1 - (6x + 1)^2}} \cdot \frac{d}{dx}(6x + 1) \] Calculating the derivative of \( 6x + 1 \), we get \( 6 \). Now substituting this into our expression: \[ y' = \frac{6}{\sqrt{1 - (6x + 1)^2}} \] Thus, the derivative is: \[ y' = \frac{6}{\sqrt{1 - (6x + 1)^2}} \]