For $y=f(x)=8 x^{3}, x=4$, and $\Delta x=0.02$ 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.

Step 1 ：Given the function \(y=f(x)=8 x^{3}\), where \(x=4\), and \(\Delta x=0.02\).

Step 2 ：To find \(\Delta y\), we calculate the difference in the function values at \(x\) and \(x + \Delta x\). This is done by substituting \(x\) and \(x + \Delta x\) into the function \(f(x)\) and then subtracting the two results.

Step 3 ：\(\Delta y = f(x + \Delta x) - f(x) = 8*(4+0.02)^3 - 8*4^3 = 7.71846399999981\)

Step 4 ：To find \(dy=f^{\prime}(x) dx\), we first find the derivative of the function \(f(x)\), which is \(f^{\prime}(x)\). Then we multiply \(f^{\prime}(x)\) by \(dx\).

Step 5 ：\(f^{\prime}(x) = 24x^2\), so \(f^{\prime}(4) = 24*4^2 = 384\)

Step 6 ：\(dy = f^{\prime}(x) dx = 384 * 0.02 = 7.68000000000000\)

Step 7 ：To find \(dy\) for the given \(x\) and \(\Delta x\) values, we substitute \(x\) and \(\Delta x\) into the expression \(f^{\prime}(x) dx\).

Step 8 ：Final Answer: a) The change in \(y\), \(\Delta y\), for the given \(x\) and \(\Delta x\) values is approximately \(\boxed{7.71846399999981}\).

Step 9 ：b) The differential \(dy=f^{\prime}(x) dx\) is \(\boxed{7.68000000000000}\).

Step 10 ：c) The differential \(dy\) for the given \(x\) and \(\Delta x\) values is also \(\boxed{7.68000000000000}\).