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If $y=f(x)=5 x^{2},$ find(a) $d y$(b) $\Delta y$(c) $\epsilon$(d) Evaluate each of these quantities if:(i) $x=10$ and $\Delta x=1;$(ii) $x=1$ and $\Delta x=0.01.$

(a) $d y=10 x d x$(b) $10 x \Delta x+5(\Delta x)^{2}$(c) $5 \Delta x$(d) $(\text { i) } 100,105,5$(ii) 0.1,0.1005,0.05

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 6

Linearization and Differentials

Derivatives

Harvey Mudd College

University of Nottingham

Idaho State University

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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03:10

01:47

Evaluate each function.

All right, so we're looking at uh Y equals X squared minus three X plus two. And they give us the the X. Value to be five and delta X to be 0.2. So the first thing I'll do is figure out what DELTA Y. Is. Uh which I can use the equation delta Y. Or delta X. Um To figure this out. Uh I guess I guess I'm kind of doing this in a weird way because if delta X is 0.2 and we find delta X as um some value minus the X. Value. So it's got to be 5.2 minus five. I guess it could have been negative but let's stick with that. Um So then what I need to do is figure out if I plug in uh five in for actual five squared is 25 minus 15 is 10 plus two is 12. And then plug in 5.2 in for X. I'm just using a calculator by the way, just typing it in. I got 13.44. Yeah. Just double checking my math. I typed it in correctly. Yeah. So then 13.44 -12 would give me 1.44 Is the Delta Y. It's only right that out. And hopefully my math made sense. Like I had a plug in 5.2 and for the excess had a plug in five for the excess and then subtract uh to get 1.44 for delta Y. Now getting delta. Uh Dy well, D Y yeah Is equal to the derivative of this which would be two X -3 times dx. Which the directions tell you that the X values five, So two times 5 -3. Yeah, Times DX which is .2. Well 2.5 -3, Sorry, two times 5 -3 is seven. Well if I do seven times 70.27 times two is 14. And I move that decimal over at 1.4. So this is my value for D. Y. And you can see How they're not off by much and that's why we can approximate using differentials because the exact value is almost as you're off by 400s. Not much of a difference at all. I believe that's the intent of this question is to compare these two numbers and see that they're pretty much the same.

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