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If $y=f(x)=3 x^{3}+1$ 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 $\Delta x=0.01;$ (iii) $x=1$ and $\Delta x=-0.01;$

(a) $9 x^{2} d x$(b) $9 x^{2} \Delta x+9 x(\Delta x)^{2}+3(\Delta x)^{3}$(c) $9 x \Delta x+3(\Delta x)^{2}$(d) (i) 900,993,93(ii) 0.09,0.090903,0.0903(iii) -0.09,-0.089103,-0.0897

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 6

Linearization and Differentials

Derivatives

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

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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All right so we're given Y equals 30 plus 12 X squared minus X cubed. They give us the X. Value and the problem would be to they give us delta X. Or or D. X. They have them equal to each other. So we can compare Is equal to 0.1. So I'm gonna solve for the delta Y. First because I can use the change in Y over change of ex formula to kind of help me out Because they tell US Delta X 0.1. Well if you think about how delta X can be written as um as X. Value minus X. Value so I'll make this 2.1 minus two is equal 2.1. I suppose you could do to minus 0.1 and your answer might be a little bit different but that's okay. So now what I'm gonna do is go back to my equation and plug in those X. Values and when I plugged into and for those exes I just used it calculated by the way, I got 70. So because I'm plugging into I got to match that up with with that. Now I'm going to plug in 2.1 into the equation again, I'm just using a calculator 73.659. So then delta y must be equal to the change in those white values. So when you do that, you'll get 3.659. Um so that means delta Y must be about 3.659. And I can only do that with a calculator. That would be very difficult to figure out 2.1 cubed and 2.1 square in the calculator and multiplied by 12. So the other piece of this is figuring out what D. Y. Is, which is A differential. We're going to find the derivative for the drift of a 30 is zero. The derivative of 12 x squared is 24 x. And then the derivative of executed is three X squared times dx. So again we get to plug in to and for these exes. So we're looking at 24 times two is 48 2 squared is four times three is 12 And then plug in 0.1 for DX. So 48 -12 is 36. And when I multiply that by .1 that just moves the decimal over My work is kind of all over the place. But anyway, dy is equal to 3.6 in the problem. So by using a differential, we're not off by much. This was 3.659 and this is 3.6. So our approximation using a differential will only be off by 59 hundreds thousands. Excuse me, 59,000 is a small decimal. So it's a good approximation if you use differentials to estimate future values

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