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# Find the derivative of the function.$g(x) = (x^2 + 1)^3 (x^2 + 2)^6$

## $\frac{d y}{d x}=6 x\left(x^{2}+1\right)^{2}\left(x^{2}+2\right)^{5}\left(3 x^{2}+4\right)$

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Samuel H.

University of Nottingham

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### Video Transcript

Let's find the derivative of G of X G of X is a product, so the first factor is X squared, plus one cubed, and the second factor is X squared, plus two to the sixth. So because it's a product, we're going to need to use the product rule to differentiate it. So the derivative will be the first times the derivative of the second plus the second time's a derivative of the first. So the first would be X squared, plus one cubed the derivative of the second. We're going to use the chain rule. On this. The outer function is the sixth power function. So you bring down the six and raise the inside to the fifth. Then you multiply by the derivative of the inside. So the derivative of X squared plus two would be two X. So what we have so far is the first times the derivative of the second plus the second time's a derivative of the first. So the second is X squared, plus two to the six and the derivative of the first. Again, we're going to need to use the chain rule. The outer function is a cubing function, so we bring down the three and we raise X squared plus one to the second. And then we multiply by the derivative of the inside and the derivative of X squared. Plus one would be to x. Okay, so we have our derivative, and now we're just going to simplify. Simplifying is the majority of the work, actually. So if we think of this as one term plus another term, let's look to see what common factors the terms have didn't that we can factor up. All right, so notice that they both have a two X so we can factor that out and notice that they both have X squared plus one this one to the third and this one to the second so we can factor x squared plus one to the second out of both, and noticed that they both have X squared plus two this one to the fifth and this one to the sixth so we can factor x squared plus two to the fifth out of both, and also noticed that we could factor three out of the six and a three out of a three so we can factor out of three. Okay, So what do we have left that we didn't factor out in the first term? We still have an X squared plus one, and we factored the three out of the six. So we still have it, too. And everything else was factored up. Now, moving on to the second term, we factored out the three. We factored out the two X, we factored out the X squared plus one squared. We factored out most of the X squared plus two to the sixth. We still have one of those left, so we have X squared plus two to the first. Okay, so now what we want to do is simplify this. So let's distribute and combine like terms. And we can also combine the three and the two and make that a six. So we have six x times X squared, plus one quantity squared, times X squared, plus two to the fifth times. Okay. What are we going to get here? We're going to have to x squared, plus another X squared. So that's going to be three x squared. And then we're going to have two times one that's two, plus another two. So that's going to be four. So here we have our simplified answer. Now, it might not be necessary for you to go through all the simplifying, but just in case, it's nice to know how to do it if you have to.

Oregon State University

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