Question

consider the following function and express the relationship between a small change in x and the corresponding change in y in the form ( dy = f(x)dx ).
( f(x)=2x^{3}-2x )
( dy=(square)dx )

Explanation:

Step1: Differentiate term - by - term

Use the power rule \((x^n)^\prime=nx^{n - 1}\).
For \(y = f(x)=2x^{3}-2x\), the derivative of \(2x^{3}\) is \(2\times3x^{2}=6x^{2}\) (since \((ax^{n})^\prime=anx^{n - 1}\) with \(a = 2\) and \(n = 3\)), and the derivative of \(-2x\) is \(-2\) (since \((ax)^\prime=a\) with \(a=-2\)).

Step2: Find \(f^\prime(x)\)

By the sum rule \((u + v)^\prime=u^\prime+v^\prime\) (where \(u = 2x^{3}\) and \(v=-2x\)), \(f^\prime(x)=(2x^{3}-2x)^\prime=(2x^{3})^\prime-(2x)^\prime\).
So \(f^\prime(x)=6x^{2}-2\).

Answer:

\((6x^{2}-2)\)