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Numerade Educator

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Problem 7 Medium Difficulty

Find the derivative of the function.
$ F(x) = (5x^6 + 2x^3)^4 $

Answer

$4\left(5 x^{6}+2 x^{3}\right)^{3} \cdot\left(30 x^{5}+6 x^{2}\right)$

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Video Transcript

let's find the derivative of this function using the chain rule. So we're going to start with the derivative of the outside function. So we have f prime of X equals. The outside function is the fourth power function. So we bring down the four and we raised the inside to the third. Okay, Now we multiply that by the derivative of the inside function, and the inside function is five x to the sixth plus two x cubed, and it's derivative would be 30 x to the fifth plus six X squared using thebe power rule. Okay, so the calculus part of this problem is done, and the rest is a bunch of algebraic simplifying and depending on your instructor, they may or may not want you to do a whole lot of simplifying, so I'll show you the amount of simplifying I would do. And then I'll show you additional steps you could do so that in the event that you see the answer looking different from yours, you can still figure out whether you have something equivalent. So I think what I would do is I would just rearrange the terms and I would write this last one after the after the 1st 1 so I would have four times the quantity 30 x to the fifth power plus six x squared, multiplied by the quantity five x to the sixth power plus two x cubed, uh, cubed. Now that's probably not the answer that you're going to see in the back of the book. Well, what I might do next is noticed that both of these terms have a common factor of six x squared. So if you factor out that six x squared and you multiplied by the four, you would have 24 x squared times five x cubed plus one. So then you write the rest of it, and that's your answer. Now that might also not be the answer that you find in the back of your book or that you're given. So what else could we dio? Well, we could notice that this term here inside the cubing function, there's a greatest common factor of X cubed. So suppose that we factor out x cubed inside the cubing function were factoring out x cubed. That leaves a factor five x cubed plus two both of those air cubed. Okay, so let's see what's next. We're going to need to get a little bit more space. Okay, so here's what we just had. And so we have X cubed, cubed, and we have five x cubed plus two cubed. So let's bring that X cubed cubed, which would be X to the ninth out and multiply it by the 24 x squared. So we have 24 x to the 11th power, and then we have our five x cubed plus one. And then we have our five x cubed plus two quantity cubed. So there's another form of the answer. And while in my opinion it might not be necessary to take it all the way to that point, it's important to know how to in case you need to.