How Do You Calculate dy and deltay for a Polynomial Function?

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SUMMARY

The discussion focuses on calculating the differential (dy) and the change in y (deltay) for the polynomial function y=2x^3-4x at x=2 with a delta-x of 0.1. The correct calculation for dy is derived using the formula dy=(dy/dx)dx, yielding a result of 2. The change in y, or deltay, is determined by evaluating the function at x+delta-x and subtracting the original function value, resulting in deltay=2.12.

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Homework Statement



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

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?
 
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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:
fitz_calc said:

Homework Statement



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

The Attempt at a Solution



dy/dt= 6(4)(.1)-4(.1) = 2[/quote
This is correct.
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)

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?
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).
 

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