# Homework Help: Differential question

1. Dec 7, 2007

### fitz_calc

1. The problem statement, all variables and given/known data

Find the value of dy and deltay for y=2x^3-4x, x=2 and dx = deltax = 0.1.

3. The attempt at a solution

dy/dt= 6(4)(.1)-4(.1) = 2

This is correct. However, I do not understand how to get deltay ? I tried 2/2*2^3-4(2) = .25 -- however, the answer is 2.12. how do they get this value?

2. Dec 7, 2007

### Defennder

Well first thing note that delta-y is simply the change in the value of y due to delta-x.

Hence delta-y= y2 - y1

where y2 is the new value of y due to delta-x (ie. y2 = 2(x+delta-x)^3 - 4(x+delta-x))

and y1 is already given as y in the equation in the problem.

Last edited: Dec 7, 2007
3. Dec 7, 2007

### HallsofIvy

Actually that is far from correct, but only because it is written incorrectly! There is NO "dy/dt" because y is not a function of t. What you meant was simply
dy= (dy/dx)dx= (6x^2- 4)= (6(4)- 4)(0.1)

y(x+ deltax)= y(2.1= 2(2.1)^3- 4(2.1)
y(x)= y(2)= 2(2^3)- 4(2)

delta y is the difference of those two: deltay= f(x+deltax)- f(x).