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Consider $f(x)=4 x^{2 / 3},$ suppose you do not have a calculator and want to approximate $f(7.9) .$ (a) Find the equation of the tangent line at $x=8$ (b) Compute the $y$ -value on the tangent line at $x=7.9$. (c) Compare your result with the answer given by your calculator.

(a) $y=\frac{4}{3} x+\frac{16}{3}$(b) 15.866667(c) 15.866387

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 2

Derivatives Rules 1

Derivatives

Campbell University

Baylor 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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Finding an Equation of a T…

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Okay, So the whole premise of this problem is approximating f of 2nd 0.9. Um, for the the function that is four x two, the two thirds power. And the whole premise of this problem is if you were to graft dysfunction e don't know exactly what it looks like. E think it looks more like this than than anything. But if you were to zoom in at X equals eight, say it's right here. Just zoom in. It looks like a straight line right there to the point where 7.9 and eight almost looks like the same value are very, very close. Eso sort of. The premise of this is that we're gonna find the tangent line, so we need the point and the slope the silk is found by taking the derivative and plugging in eight into the problem that we have been. Write that down f prime of eight on. Then we can come up with the tangent line and plug in 7.9 and see what we get. So the first thing is let's plug in eight and for the function, The cube root of eight is two squared is four times this four. Give me 16. And then, uh, what we also have to do is figure out what f prime is. Well, if you bring that two thirds in front, you get eight thirds and then next to the negative one third power well, again, we need to plug in eight into that. Well, the cube root of eight is too. And you could take a divide by to give you four thirds eso. Why would we go through that process is now we can say that the equation of the tangible line wife was four thirds ex Manage your X coordinates plus 16. Now, this doesn't seem that helpful. Except now what you can do is plug in 7.9 into this problem and we can approximate four thirds. Let's see, 7.9 minus eight would be negative. 1/10. They have 1/10 times four thirds and add 16 to that. Really, that's often bid. Now that's pretty close on, baby. Even reduce that both those individual by two. So looking at negative 23 times five is 15 plus 16. A good approximation to that value is about on 15.8666 kind of weird that they say without a calculator, and they give you an answer with a calculator. But when you actually compare that to the actual answer Ah, 15.8663 we were good for what? The nearest 1000th. We weren't off until we went to the nearest 10,000. So it's a pretty good approximation. I think that's what they want you to get by. When you look at part C in the problem, Here's part P. I don't label. By the way, this is part a so there you have it.

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