Let $y=2 x^{2}$. Find the change in $y, \Delta y$ when $x=5$ and $\Delta x=0.3$ Find the differential $d y$ when $x=5$ and $d x=0.3$

Step 1 ：Let \(y=2 x^{2}\).

Step 2 ：Find the change in \(y\), \(\Delta y\), when \(x=5\) and \(\Delta x=0.3\).

Step 3 ：Find the differential \(d y\) when \(x=5\) and \(d x=0.3\).

Step 4 ：The change in \(y\), \(\Delta y\), is the difference in \(y\) values when \(x\) changes by \(\Delta x\). So we need to calculate \(y\) at \(x=5\) and \(x=5.3\) and find the difference.

Step 5 ：The differential \(d y\) is the derivative of \(y\) with respect to \(x\), multiplied by the change in \(x\), \(d x\). So we need to find the derivative of \(y\) with respect to \(x\) and multiply it by \(d x\).

Step 6 ：Calculate \(y\) at \(x=5\) and \(x=5.3\): \(y_5 = 50\), \(y_{5.3} = 56.18\).

Step 7 ：Find the difference: \(\Delta y = y_{5.3} - y_5 = 6.18\).

Step 8 ：Find the derivative of \(y\) with respect to \(x\): \(dy/dx = 4x\).

Step 9 ：Calculate \(d y\) when \(x=5\) and \(d x=0.3\): \(d y = dy/dx * dx = 6\).

Step 10 ：Final Answer: The change in \(y\), \(\Delta y\), when \(x=5\) and \(\Delta x=0.3\) is \(\boxed{6.18}\). The differential \(d y\) when \(x=5\) and \(d x=0.3\) is \(\boxed{6}\).