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If $y=f(x)=\left(x^{3}+5\right)^{2}$ find(a) $d y;$(b) $\Delta y;$(c) $\epsilon$;

(a) $6 x^{2}\left(x^{3}+5\right) d x$(b) $6 x^{2}\left(x^{3}+5\right) \Delta x+\left(30 x+15 x^{4}\right)(\Delta x)^{2}+20 x^{3}(\Delta x)^{3}+6 x(\Delta x)^{5}$$+10(\Delta x)^{3}+(\Delta x)^{6}+15 x^{2}(\Delta x)^{4}$(c) $\left(30 x+15 x^{4}\right)(\Delta x)+20 x^{3}(\Delta x)^{2}+6 x(\Delta x)^{4}+10(\Delta x)^{2}$$+(\Delta x)^{5}+15 x^{2}(\Delta x)^{3}$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 6

Linearization and Differentials

Derivatives

Campbell University

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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for this problem we are given Y equals three X squared plus five x minus two. And in part A we are asked to find delta Y. So we know that delta Y equals f of X plus delta x minus f of X. So this will be equal to and I'll write this out first, I guess we'll have three times X plus delta X squared plus five X plus five delta X minus two. Then we'd have minus three X squared Plus. or excuse me not plus -5 x plus two. So plus two minus two at +20 minus five X plus five, X add to zero off the bat. Then expanding out our brackets there, we'll get that this becomes three X squared plus six delta X or six X delta X. I should say X delta X plus three delta X squared Plus five Delta X minus three X squared three X squared is add to zero. So we're just left with six X delta X plus. Oh actually we can simplify this further. We can write this as six X plus five delta X plus three delta X squared. Then, for part B finding dy we know that that is equal to f prime of X dx And we should be able to easily see that f prime of X is simply going to be six X plus five. So this becomes six X plus five times dx. And for part C we're asked to find dy minus delta Y. So that's going to be equal to six X plus five. Delta X -6 x plus five delta X minus three delta X squared. So dy minus delta Y is going to be equal to negative negative three delta X squared.

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