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Use logarithmic differentiation to find the derivative.$$y=\frac{(3 x-2)^{8} \sqrt{4 x-1}}{x^{2}+1}$$

$$\left(\frac{24}{3 x-2}+\frac{2}{4 x-1}-\frac{2 x}{x^{2}+1}\right)\left(\frac{(3 x-2)^{8} \sqrt{4 x-1}}{x^{2}+1}\right)$$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 6

Properties of Logarithmic Functions

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Use logarithmic differenti…

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find the derivative using …

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Use logarithmic diff…

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eso The key to this problem, as I copy things down is ah. When you take the derivative of such an ugly thing like this, what you want to do is rewrite things. Um, by taking the log of both sides e have to rewrite this problem and getting shortly so hopefully I can squeeze this on one piece of paper. So the key thing is that why equals this? So remember all of this that's in black, because I have to rewrite that part again. Um, so just for me showing my work, I mean natural log both sides. There is why you want a natural log both sides is. Then you can split up into three different logs. Um, now, instead of me writing natural log of three X minus two to the eighth power, I'm also gonna move that eight in front and then plus, And so me just writing natural law before X minus one to the one half power because of the square root. Uh, I'm gonna bring that one half in front. And then for the division, you subtract natural log of X squared plus one. So the left side is still that natural log of of Why? So now when I take the derivative of the left side, it's gonna be one over Y d Y d x. Okay. And then everything else I have eight over. Uh, the director of the natural log is three x minus two goes in the numerator and then times by the derivative of the inside, it is the chain rule. It s the same thing with the next 11 half times. Um, that four x minus one goes into the numerator and then times and a review of the inside is times four and then same thing here. Wow, Natural logs. Derivative is one over X squared, plus one times the derivative of the inside is two X. Now, I will clean this up a little bit. But what I also need to do is solve for D y d X by multiplying this wide over. So what? You'll notice that part of my answer if I'm getting do y The X by itself, is I also have to multiply by Why? Which is the original problem I had written in black earlier. So that needs to replace for this way down here, which is that I was trying to squeeze it. That three X minus two to the eighth power the square root. Ah, four X minus one and over. X squared plus one. Um So what I'm doing is solving for D Y d X is part. That answer is part of my answer. Um, noticed my use of parentheses, too. I'm looking that half of force to so to over four X minus one. And I had to clean up this one. And there's your answer pretty long. But it is correct, Yeah.

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