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Find $d y / d x$ if $y=\sqrt{\frac{x+1}{x-1}}$.

$$\frac{-1}{(x-1)^{3 / 2} \sqrt{x+1}}$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 6

The Chain Rule

Derivatives

Campbell University

Harvey Mudd College

Baylor 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.

02:29

find $d y / d x$$y=\fr…

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Find $d y / d x$$$y=\s…

00:47

Find $d y / d x$.$$y=\…

01:05

Find $d y$$$y=\sqrt{x+…

00:54

Find $\frac{d y}{d x}$.

03:48

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04:14

Find $d y / d x$.

…

01:32

Find $d y / d x$. $$y=\sqr…

01:42

Find $d y / d x$.$$y=x…

02:13

Find $d y / d x$. $$y=\fra…

00:53

Find $d y / d x$.$$

he said. So we're gonna find the derivative using the chain and the quotient rule of this giant square root. We have a X plus one in the numerator and X minus one in the denominator. So what I would do is rewrite this problem. Still the quotient. But now to the one half hour eso as I take the derivative of this, the first thing to do would be to take the one half in front. You leave that question alone is now to the negative one half power. And you have that X plus one over X minus one, and then you multiply by. Well, the inter function is going to be the quotient rule where you do the derivative of the top is one. You leave the bottom alone, minus you take the derivative of the bottom, which is one you leave the top alone all over the denominator squared X minus one squared. Now, some teachers will actually let you leave your answer like this. Other teachers, not so much. What they'll have you do is this thing about Oh, well, if we have a negative exposure like we do here, we have to flip this because it's a reciprocal and then square root each piece so that's looking at the square root of X minus one over the square root of X plus one. Because, remember, you gotta flip it. One half is still there, and then here. If you wanted to simplify, just distribute that negative. Well, one X minus one x zero x negative one minus one Would be, uh, so times negative to there over X minus one squared Notice. I'm making this a giant fraction because we multiply straight across. If you wanted to, you could just cancel out some twos. Remember, this is to the one half power so you can subtract exponents, and probably a better answer would be if you go, you know, we still have a negative in there and would be X minus one after you subtract your exponents would be to the three has power and then X plus one would be to the one half hour. This is probably how most math teachers would write it, but some teachers will let you leave it like this. Others like this

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