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?

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.

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

It's understandable that you're feeling overwhelmed since you're trying to learn this on your own. Don't worry, I'm here to help!

Let's break down the steps and clarify the concepts involved in this problem. First, we need to differentiate the function y = 2 - x^4 with respect to x, which you correctly did, getting dy = -4x^3.

Next, we can evaluate dy at x = 2 by substituting the value into the expression we obtained for dy. So, dy = -4(2)^3 = -4(8) = -32. Therefore, the value of dy when x = 2 is -32.

Now, let's move on to delta y. Delta y represents the change in y, which we can calculate using the given value of delta x. To find delta y, we need to evaluate the function f(x) = 2 - x^4 at x = 2 + 0.01 and subtract the function value at x = 2.

Plugging in x = 2 + 0.01 into the function, we get f(2 + 0.01) = 2 - (2 + 0.01)^4 = 2 - 2.000001^4. After evaluating this, we find that f(2 + 0.01) ≈ 1.99839.

Next, we need to find f(2). Plugging x = 2 into the function gives f(2) = 2 - 2^4 = 2 - 16 = -14.

Now, we can calculate delta y by subtracting f(2) from f(2 + 0.01), that is, delta y = f(2 + 0.01) - f(2) = 1.99839 - (-14) = 15.99839. Therefore, delta y is approximately 15.99839.

To compare the values of dy and delta y, we can say that dy represents the instantaneous rate of change (slope) at a specific point (x = 2), while delta y represents the change in y over a small interval (delta x = 0.01). In this case, dy = -32 and delta y ≈ 15.99839.

Thus, we can conclude that dy and delta y have different values because they represent different quantities: dy represents the rate of change at a single point, whereas delta y represents the change in y over a small interval.

I hope this explanation clarifies things for you! If you have any further questions, feel free to ask.