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This exercise uses (6) to prove the chain rule. Suppose $y=f(u)$ and $u=g(x)$ are each differentiable functions.(a) Show that $$\Delta y=f^{\prime}(u) \Delta u+\epsilon_{1} \Delta u$$where $\epsilon_{1}$ approaches zero as $\Delta u$ approaches zero.(b) Show that $$\Delta u=g^{\prime}(x) \Delta x+\epsilon_{2} \Delta x$$,where $\epsilon_{2}$ approaches zero as $\Delta x$ approaches zero.(c) Substitute (b) into (a) to show that $$\Delta y=f^{\prime}(u) g^{\prime}(x) \Delta x+\epsilon \Delta x$$,where $\epsilon=\epsilon_{1} g^{\prime}(x)+\epsilon_{2} f^{\prime}(u)+\epsilon_{1} \epsilon_{2}$(d) Show that $\epsilon$ approaches zero as $\Delta x$ approaches zero. Explain why (c) and (d) together prove the chain rule.

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 6

Linearization and Differentials

Derivatives

Missouri State University

Baylor University

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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were given a function were asked to show its differential at every point. There are a few different ways to do this. We could use the serum from the book that says that if the function it's partial, derivatives are continuous at the point and it's differential. But the point, however, there's another way of the book is asking us to do so. Consider Delta Z. This is the same as half of explosives. Delta X Why plus Delta y minus f of x y. We have that This is equal to explodes still. Two x cubed, plus three times y plus Delta y minus X cubed minus three y, which is equal to X cubed minus X cube. Those cancel out and three x squared time still two x plus three x Delta X squared plus built X cute plus three y minus three. Why, that cancels out Plus three dealt Why you want to write this in a form with differentials? So, first of all, notice that effects of X Y is three x squared and if why of X Y is equal to three. So we have that this is going equal to three x squared. You recognize this is FX X one Delta X plus, and then three. Delta y Recognize is that's why X Y don't know why. And then we also have this extra term here, which is three x Delta X Square plus Delta X huge. And so let's call Epsilon one. This is three x Delta X plus. Does that fix squared? This clearly approaches zero as Delta X and Delta y purchase zero zero, because this is the same as the limit has simply felt exit perching zero. And we have that Absalon too. Well, this is simply going to be zero. Since we have no extra Delta y terms, this clearly is going to approach zero as Delta X Delta y approaches 00 And so we have shown that because both of these terms, that's a Mormon Epsilon to approach zero we have that has Delta, X and Delta y zero Delta Z is going to approach or perhaps better, will you, right? This is yeah, we're going to see it. Yes, we have that. The function if the X y is differential and since in your calculation X and Y are arbitrary, this is differential everywhere

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