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Catherine R.

Missouri State University

Kristen K.

University of Michigan - Ann Arbor

Samuel H.

University of Nottingham

Michael J.

Idaho State University

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Video Transcript

eso I would rewrite this problem as X tangent of X to the Nega first power. And that is one over X. Um, that just might help you. When you get to the chain rule, you have your inter function and your outer function and right here is telling you to do the product rule. So the first derivative you take the drift of of X, which is one you know how to write that you leave the tangent of one over x alone. Plus now you leave X alone, the derivative of tangent seeking squared and you leave the one over X alone or extra the negative one. And then you have to multiply by the drift of of this, which is negative one ext of a negative second power. Now, before I do anything from here, I would just simplified this a little bit because if you add your exponents, um, you end up with X and the denominator, and I should have rewrote that as minus because you have a negative one there. Um, yeah, one of those excess will cancel. So now from here, I can't find the second derivative because that's what the direction I want you to do. And this is your chain rule, just like what we did earlier. Except you don't have the X in front. Um, so the drift of tanginess seeking squared of one over X. Um and you already see I wrote the negative here because this is the same thing Is before accident a get first. So negative one exiting of second over X squared. So now this is a different kind of rule. Um, because it's a quotient rule and the drift of of the top is a chain rule for sure. Where you want to bring the two in front, You leave seeking of one over x alone. But then the derivative of seeking is seeking tangent. I mean, right Siegen squares on to write that again, tangent of one over X and then the derivative of one of her ex. Just like what I had before is that negative? One Excellent. And of second power times have on next hour. And then you need to leave the denominator alone, which is times X minus the drift of of the bottom which is just one. You leave the top alone. You can squared of one over X all over the denominator squared. Now I think I'm past my time of simplification, but there's a lot of things that you can simplify. I can put these fractions together because they're both over X squared. This piece might put a hiccup in that you could factor things out. There's so much more you can do. I don't know where to stop, So I'm gonna stop right there.

Catherine R.

Missouri State University

Kristen K.

University of Michigan - Ann Arbor

Samuel H.

University of Nottingham

Michael J.

Idaho State University

Lectures

Join Bootcamp