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

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We start by letting $$ u = 6x+1 $$, then the function becomes 
$$
y = \sin^{-1}(u).
$$

Using the chain rule, the derivative of $$ y $$ with respect to $$ x $$ is
$$
y' = \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}.
$$

The derivative of $$ u $$ is
$$
\frac{du}{dx} = 6.
$$

Substituting back in, we obtain
$$
y' = \frac{6}{\sqrt{1-(6x+1)^2}}.
$$

Thus, the derivative is
$$
y' = \boxed{\frac{6}{\sqrt{1-(6x+1)^2}}}.
$$
The derivative of $$ y = \sin^{-1}(6x+1) $$ is $$ y' = \frac{6}{\sqrt{1-(6x+1)^2}} $$.

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

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