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?
Hold on; I think I did delta y wrong. If you set it up like that, then you would get 2-(4.01)^4 -2+2^4. And you get a ridiculous number...?
Wait, never mind. I'm overthinking it and I made a dumb mistake. Disregard these two answers.
I understand that you're struggling with the concept and how to approach the problem. Let me break it down step-by-step and explain the reasoning behind each step.
First, you correctly differentiated the function y = 2 - x^4 to find dy/dx = -4x^3. This is the derivative of the function, which gives you the rate of change of y with respect to x.
To evaluate the two quantities, delta y and dy, you need to substitute the given values into the derivative equation. Let's work through it:
1. For dy:
Substituting x = 2 into dy/dx = -4x^3 gives you dy = -4(2)^3 = -32.
2. For delta y:
Here, you need to find the change in y when x changes by a small amount, dx. In this case, dx = 0.01.
So, you calculate f(2 + 0.01) - f(2), where f(x) represents the original function, y = 2 - x^4.
f(2 + 0.01) = 2 - (2.01)^4
f(2) = 2 - (2)^4
Evaluating these expressions, you get:
f(2 + 0.01) - f(2) ≈ -0.3224
Based on your calculations, it seems you've arrived at similar results. Your value for dy (-0.32) is approximately the same as -32, and your value for delta y (-0.3224) aligns with our approximation of -0.3224.
To compare these values using words, you can say that dy represents the instantaneous rate of change of y at a specific point (x=2). It measures the slope of the tangent line to the function at that point. On the other hand, delta y represents the change in y over a small interval (dx = 0.01) around x=2.
Comparing the absolute values of dy and delta y, you can see that delta y is smaller than dy. This implies that the change in y is smaller over a small interval (delta y) compared to the instantaneous rate of change (dy) at that specific point.