For $y=f(x)=6 x^{3}, x=2$, and $\Delta x=0.06$ find a) $\Delta y$ for the given $x$ and $\Delta x$ values, b) $d y=f^{\prime}(x) d x$, c) $d y$ for the given $x$ and $\Delta x$ values. a) $\Delta y=$ (Round to four decimal places as needed.)

Step 1 ：Given the function \(y = f(x) = 6x^3\), the value of \(x = 2\), and \(\Delta x = 0.06\).

Step 2 ：To find \(\Delta y\), we need to calculate the difference in the function's value at \(x\) and \(x + \Delta x\). In other words, we need to calculate \(f(x + \Delta x) - f(x)\).

Step 3 ：Substitute the given values into the equation, we get \(\Delta y = f(2 + 0.06) - f(2)\).

Step 4 ：After calculating, we find that \(\Delta y = 4.45089600000000\).

Step 5 ：Rounding to four decimal places, we get \(\Delta y = 4.4509\).

Step 6 ：Final Answer: For part a), \(\Delta y\) is approximately \(\boxed{4.4509}\).