Question

Compare the values of \(d y\) and \(\Delta y\). \(y=1-2 x^{2} \quad x=0 \quad \Delta x=d x=-0.1\)

Short answer

The calculated values are \(dy=0.0\) and \(\Delta y=-0.02\).

Step-by-step solution

Calculate the derivative of the function.

Differentiate the function \(y=1-2x^{2}\) to find \(y'\) or \(dy/dx\). By applying the power rule, the derivative is \(-4x\).

Calculate dy

Using the derivative, \(dy= y'(dx)\). So, using the derivative found in step 1 and the given \(dx=-0.1\), \(dy= -4(0)(-0.1)= 0.0\).

Calculate Δy

To compute Δy, evaluate the given function at \(x + \Delta x\), and at \(x\), then subtract these two results. If we substitute \( x + \Delta x = 0 - 0.1 = -0.1 \) into the function, we get \(y_1 = 1 - 2(-0.1)^2 = 0.98\). Now, for \(x = 0\), the function value or \(y_2\) is 1. So, Δy is \(y_1 - y_2 = 0.98 - 1 = -0.02\)