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If $f^{\prime}(u)=u^{2}-1,$ and $y=f\left(x^{2}\right),$ find $d y / d x$.

$$2 x\left(x^{4}-1\right)$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 6

The Chain Rule

Derivatives

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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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I think this problem is written mawr confusing than it should be. Uh, like I think there's a better way of writing. This is my point. Eso Let's start with the second statement why it was f of x squared. Because then if I take the derivative of that, we're looking at the chain role where what you do is the derivative of F, but then you have to multiply by the directive of the inside. Well, then, from here, what I would do is look at that other statement where it says f prime of you, Uh, like you find it is equal to use squared minus one. So what that tells me is that I can replace this piece. I'll go back toe black, So, uh, f prime is defined as something squared minus one, but we'll bracket around here. We still have this two X right here. But what I need to do is replace uh, x squared in for you right here. So the way I would write the answer then is first of all, I moved to X in front, just out of habit. Most people do that and x squared squared. Well, you multiply the exponents that becomes X to the fourth minus one. And that's what I would do to get this answer. It makes more sense to me this way, then kind of how they wrote the question. But anyway, here is what we need.

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