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# Find the derivative of the function.$y = [x + (x + \sin^2 x)^3]^4$

## 4$\left[x+\left(x+\sin ^{2} x\right)^{3}\right]^{3}\left[1+3\left(x+\sin ^{2} x\right)^{2}(1+2 \sin x \cos x)\right]$

Derivatives

Differentiation

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### Video Transcript

Here's a composite function that we're going to differentiate using the chain rule and the good news is that even though there are going to be a lot of steps because we have a lot of layers in this composite function, there's not going to be any simplifying. Weaken do and we're done. So that is often a relief because the simplifying is usually the most complicated part of the process. So let's find the derivative notice that the outermost function is the fourth power function. So we bring down the four and we raised the inside to the third. So we just write the entire inside over again. But now it's to the third. Now we multiply by the derivative of the inside. So just focusing on this part, which is a some and the derivative of the sum, is going to be the some of the derivatives. So the derivative of X is one plus. Now we're finding the derivative of this inner part, and we're going to need to use the chain rule on that. So we bring down the three and we raised the inside to the second. Then we multiply by the derivative of the inside, and the inside is a some. This part here is a some and so it's derivative will be the some of the derivatives. The derivative of X is one plus. Now we need the derivative of sine squared X, and remember that sine squared X means the sign of X quantity squared. And so it's derivative would be, according to the chain rule, two times a sign of ex because you bring down the two and you raise sign to the first multiplied by co sign of X because that's the derivative of sign. So that's what we have next to sign X Co sign X. Okay, we made it to the very inside and then we look at that for a minute and we ask ourselves, What can we combine? What can we factor out? And we realized that because of the plus signs, primarily it's kind of stuck the way it is. So there you have it

Oregon State University

Derivatives

Differentiation

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