Step 1 :Given the function \(y = e^{3x} + 3x\), we are asked to compute the values of \(dy\) and \(\Delta y\) at \(x = 0\) with \(\Delta x = dx = 0.05\).
Step 2 :The differential \(dy\) is given by the derivative of the function times \(dx\), i.e., \(dy = f'(x)dx\).
Step 3 :The increment \(\Delta y\) is given by the difference in the function's values at \(x + \Delta x\) and \(x\), i.e., \(\Delta y = f(x + \Delta x) - f(x)\).
Step 4 :First, we need to find the derivative of the function \(f(x) = e^{3x} + 3x\). The derivative is \(f'(x) = 3e^{3x} + 3\).
Step 5 :Substitute \(x = 0\) and \(dx = 0.05\) into the expression for \(dy\) to find its value. The result is \(dy = 0.3\).
Step 6 :Substitute \(x = 0\) and \(\Delta x = 0.05\) into the expression for \(\Delta y\) to find its value. The result is \(\Delta y = 0.3118\).
Step 7 :Final Answer: \[\begin{array}{l}d y=\boxed{0.3} \\Delta y=\boxed{0.3118}\end{array}\]