Determine $$ \frac{d y}{d x} $$ for $$ y=\cos ^{-1}\left(2 x^{6}\right) $$

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We start with  
$$
y = \cos^{-1}\left(2x^6\right).
$$

1. Differentiate using the chain rule. The derivative of $$\cos^{-1}(u)$$ with respect to $$u$$ is  
$$
\frac{d}{du} \left[\cos^{-1}(u)\right] = -\frac{1}{\sqrt{1 - u^2}},
$$  
and here $$u = 2x^6$$.

2. Compute the derivative of $$u = 2x^6$$ with respect to $$x$$:  
$$
\frac{du}{dx} = 12x^5.
$$

3. Applying the chain rule, we have  
$$
\frac{dy}{dx} = -\frac{1}{\sqrt{1 - \left(2x^6\right)^2}} \cdot 12x^5.
$$

4. Simplify the expression under the square root:  
$$
\left(2x^6\right)^2 = 4x^{12},
$$  
so the derivative becomes  
$$
\frac{dy}{dx} = -\frac{12x^5}{\sqrt{1 - 4x^{12}}}.
$$

Thus, the final answer is  
$$
\frac{dy}{dx} = -\frac{12x^5}{\sqrt{1 - 4x^{12}}}.
$$
$$ \frac{dy}{dx} = -\frac{12x^5}{\sqrt{1 - 4x^{12}}} $$

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