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Use logarithmic differentiation to find the derivative of the function.

$ y = \sqrt x e^{x^2 -x}(x + 1)^{2/3} $

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01:05

Frank Lin

01:49

Doruk Isik

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 6

Derivatives of Logarithmic Functions

Derivatives

Differentiation

Missouri State 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.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

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Use logarithmic differenti…

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Use logarithmic diff…

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find the derivative using …

So if we want to take the derivative of this, we can imply the natural log on each side to kind of help us out a little bit. Because first, if we go ahead and rewrite this, so remember, this is actually X to the one half power. Thanks. So any time we have things multiplied together, we can write it as the sum of the logarithms. And then if we have any powers, we can move those out front. So we could actually rewrite this in the following way being so this route X becomes one half natural log of X and then plus, um well, the natural log of E just cancels all we're left with the power. So would be X squared minus x. And then if we take the natural log of, um, this over here, that would be plus two thirds natural log of X plus one. Okay, Now we can go ahead and take the derivative on each side, so just apply the derivative, okay? And now on the left, we're going to have to use chain rules or the derivative of natural log. Um, is going to be first one over. Why, but Then we take the derivative of the inside, which would just be why prime? Now we come over here and take the derivative of so it would be one half and then natural log of access. Derivative is just one of our X, um the derivative of X squared. We use power rules that would be plus two X. The derivative of X is going to be once that would be minus one and then we have the derivative of two thirds. So we take the derivative of natural log exports. One supposed to be one of our exports one and then we take the derivative of the inside. So the derivative of X plus one is just one. And now all we need to do is multiply over by one. So we get why prime is equal to why times So the 1/2 x plus two x minus one plus, uh, two thirds X plus one. And now, at this point, we can just go ahead and replace why, with what are starting function was I mean, or you can really leave it like this if you want, but I'm just gonna go ahead and plug it in, um, so it would be why Prime is equal to so it be Route X E to the X squared minus x X plus one times two thirds. And then times what we had before. So 1/2 x plus two X minus one plus 2/3 X plus one. So, again, either of these are valid ways of writing. The solution just kind of depends on which way you kind of need to write it. Um, which way is kind of more convenient? Um, yeah. Normally, I would write as the second way, but a lot of times you can just get away by just writing in the first way as well, at least depending on what you're trying to accomplish.

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