Question

Please, don't assume that we've been taught this stuff because we were literally just handed a packet and told to learn it ourselves. I don't understand this in the slightest.

The question is: Use the information to evaluate and compare delta y and dy. y=2-x^4, x=2 and delta x = dy= 0.01

What I've done so far is differentiate, to get that dy=-4x^3. In the packet, it looks like they plugged 2 in for x and multiplied by dx, so I did, and I got -.32

For delta y, I set up the equation like f(2+0.01) - f(2), and I got -.3224

Are these right? Is there some way that I'm supposed to compare them using words? If I don't even understand what I'm doing, then how am I supposed to do so?

Answer 1

What I can do is to lead you to some pages that explain the difference in
dy and ∆y
I thought this youtube did a good job
http://www.youtube.com/watch?v=xoAnLgs6hiM

There are also numerous other youtube links listed in the right column

btw, I agree with your answers.

Answer 2

Delta y is the Actual change of y when x changes from x to x+delta x. (I capitalized Actual so that you'd see the top part of A looks like delta - a way to remember what delta y is.) dy is the estimated change in y when x changes from x to x+delta x when you use the tangent line at x to estimate.

To help you see this, let's look at the Actual change of y, or delta y.

x starts at 2 and moves 0.01 to x=2.01
f(2)=2-2^4=-14
f(2.01)=2-2.01^4=-14.32240801
Therefore the change of y, delta y = -0.32240801

The equation of the tangent line at x=2 is y=-32x+50. (This requires work on the "side.")

If x=2, y=-64+50+-14
If x=2.01, y = -64.32+50=-14.32
Therefore dy=-.32

But that's too much work!
Remember calc notation:
dy/dx=f'(x)
"Multiplying" that equation by dx gives
dy=f'(x)*dx
This is a faster way to find dy!

f(x)=2-x^4
f'(x)=-4x^3
At x=2,
dy=(-4*2^3)*(.01)=-.32

Notice this is what we got with the tangent line approximation which is very close to the "Actual" change of y or delta y using the "Actual" function!



LON