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Problem 4

In Exercises $1-20,$ find the derivative. $$y=\l…


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Problem 3

In Exercises $1-20,$ find the derivative.
$$y=(\ln x)^{2}$$


$y^{\prime}=\frac{2 \ln x}{x}$



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Video Transcript

it's so we're continuing to work through some of the examples in 3.9 that incorporate looking at derivatives of exponential zenn la girth MK functions. So again, just a quick recap. You need to be very familiar with your basic differentiation rules for Constance additions of traction products quotients, chain and power because thes air gonna get used regardless of what kind of function we're dealing with to begin with. And then we just incorporate how the derivatives occur with the new functions themselves. So looking over the new ones the exponential logarithmic functions that came up in this chapter you've got your drivel of the E function, which is still e. You've got the derivative of a power with a variable exponents that becomes the power itself times the natural log of your base. You have the derivative of a natural log that becomes one over whatever the natural log is. And then, if that's a function, we have to apply the chain. Ruelas Well, so we take the derivative of the one and then multiply it by the derivative of the new function. Same with your log. So the drivel of a basic log function becomes one over X times, the natural log of whatever the base waas and again, if that natural log of your base and then applying your chain rules So taking the derivative of that function itself. So let's look at how the supplies to this question. So I have a natural log of X and a squared, so I'm dealing with my power rule. So my basic power rule, which is anything with a variable power, becomes that power gets moved in front. Our base stays the same, and our power becomes one less. But because I also have a function in the brackets, I'm going to be dealing with my chain rule. So I have to deal with my power up front, and then I have to deal with my function inside. And I'm gonna run out of room for this so we'll just see if I can kind of squeeze this one in. We'll just move that over chain role. So again, you've got multiple functions being put together, deal with your derivative of your one, and then do the derivative of your inside and then we're applying our natural log derivatives. So we're also applying the one that we learned in this chapter which the derivative of a natural log becomes one over X. So applying that here to this question would look like the following. So we moved the two in front. We deal with that. First, leave our function and brackets the same, and our power becomes one less. Now we apply the chain rule and we go to the inside of the brackets, which is our natural log of X, and we do that derivative as well. From there, we can kind of clean up the process because there's not much more I could do. So I would have my two element of x two minus one is one. And then I'd have my one over X function as well, so you could leave it like that. There's a bunch of other ways you can put it. I mean, you could put your ex with the two. Write that with your Ellen of ex. You could write that as to Elena Vex all over X, or you could even leave it as the top one. They all mean the same. You don't actually have to write this little one right here because anything to the power of one is one. And depending on how you look at it, the textbook could have it listed as one way or the other. They both mean the same thing, though.

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