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Find $ f'(a) $.

$ f(x) = \sqrt{1 - 2x} $

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04:31

Daniel Jaimes

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 7

Derivatives and Rates of Change

Limits

Derivatives

Missouri State University

Campbell University

University of Nottingham

Idaho State University

Lectures

04:40

In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.

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.

0:00

03:51

Find $$f^{\prime}(a)$$…

09:28

01:23

Find $f$ such that:$$f…

02:01

Find $f^{\prime \prime}(x)…

02:54

03:59

find $f^{\prime}(x)$$$…

01:50

Find $f^{\prime}(x)$

Okay. And this function F of X is equal to the square root of one minus two X. And we want to find the derivative at a. We are going to use the chain role which says the derivative of our function with respect to X. Dy dx equals the derivative of the function with respect to you times the derivative of U. With respect to X. And you in this case is going to be the one minus two X. So we're taking the square root of one minus two X. Or the square root of you. So now I'm going to rewrite our function F. Of X As 1 -2. X. And instead of writing as the square root of 1 -2 x. uh the square root is the same thing as raising to the one half power. So F of X can be rewritten as one minus two X. To the one half power. Now this is F. Of X. We're now going to differentiate dysfunction. Now if you think of this as the U. The derivative of U. To the one half is one half times you. Which is this in parentheses. Uh And then one half, You would have to subtract one from this exponential. You would get one half minus one is negative one half. So that's the derivative of You to the 1/2. It's one half times U to the negative one half. But then it's little hidden here. Uh We still have to do the derivative of U with respect to X. Let me show you that. Okay, we're using the chain rule. Uh to find F. Prime of X. We had to do D. Y. D. U. We just did that part. The derivative of U. To the one half is one half times U. To the negative one half. But then we had to finish by multiplying by the derivative of U. With respect to X. Remember this thing into parentheses to one minus two X. Is our youth. So D. U. D. X. The derivative of our function you with respect to X is simply the derivative of one minus two X. With respect to X. Which is negative two. So this whole thing is going to times negative two. And you can see that the one half times a negative two will give you negative one. And the one minus two X. To the negative one half is the same thing as one minus two X. To deposit of one half in the denominator. Uh And of course one minus two X. Uh to deposit of one half is really the square root Of 1 -2 x. So f prime of X is equal to this. And uh so F prime of a the derivative of F evaluated at a simply plug a infrared X. We're gonna have negative one over. uh the square root of 1 -2 times a

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