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Find the derivative of the function.$ f(x) = (2x - 3)^4 (x^2 + x + 1)^5 $

$f(x)=(2 x-3)^{3}\left(x^{2}+x+1\right)^{4}\left[28 x^{2}-12 x-7\right]$

00:34

Frank L.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Derivatives

Differentiation

Campbell University

Harvey Mudd College

University of Michigan - Ann Arbor

Boston College

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Let's find the derivative of F of X and F of X is a product. The first factor is two X minus three to the fourth on. The second factor is X squared plus X plus one to the fifth. So we're going to use the product rule. So we start with the 1st 2 X minus three to the fourth times, the derivative of the second and the second is a composite. So we're going to need to use the chain rule. So the derivative of the outside the outside would be the fifth power function, So we would bring down the five and raise the inside to the fourth. So that takes care of the derivative of the outside. Now we multiply it by the derivative of the inside, the derivative of X squared plus X plus one would be to X plus one. So what we've done so far is the first times the derivative of the second. Now we're gonna do plus the second X squared plus X plus one to the fifth times. The derivative of the first and the first is also a composite. So we're going to use the chain rule to find the derivative of two X minus three to the fourth. The outside function is the fourth power function. So we bring down the four and we raised two X minus three to the third. And now we multiply by the derivative of the inside. The derivative of two X minus three is two. Okay, so we have our derivative, and now it's a matter of simplifying. So this entire first part is our first term. And this entire second part is our second term. Squeeze my one back in there. Let's see if we have any common factors that weaken factor out of both terms. It looks like both terms have two X minus three. This one has it to the fourth. This one has it to the third so we can factor out two X minus three to the third from both of them. And it looks like both terms have X squared plus X plus one. This one has it to the fourth. This one has it to the fifth so we can factor out X squared plus X plus one to the fourth from both of them. So then the remaining factor, what we still have in the first part. We still have a two X minus three. We still have a five and we to still have a two X plus one. So we'll put the five first five times two x minus three times two x plus one plus. Now let's look at what we have left in the second factor the second term. So we factored out four of these and we still have one left. So we still have X squared plus X plus one. We still have the four. We still have the two, and we factored all of those. So four times, too. We have eight times X squared plus X plus one. Now it may not be important or necessary for you to do all of this simplifying. It just depends on what your instructor requires, but it's good to know how, just in case you need to. Okay, the last thing I'm going to do is simplify this and I'm going to do that by using the foil method on the binomial, distributing the five and then combining like terms distributing the eight as well in combining like terms. So we're going to have are two x minus three cubed our X squared plus X plus one to the fourth. And then okay, when we foil, we get four x squared and we're multiplying that by five. So 20 x squared we get plus two x and minus six x So that's minus four X. And we're multiplying that by five. So minus 20 x and we get minus three and we're multiplying that by five. So minus 15 we distribute the eight we get eight x squared eight x and eight. Okay, the last thing we need to do is combine the like terms. Gonna give ourselves a little bit more room to work here. And here's what we just had on the last step. So we're going to combine 20 x squared and eight x squared, and we have 28 x squared. We're going to combine negative 20 x and positive. Eight x. We get negative 12 x and we have negative 15 and positive eight. So that would be minus seven. So that times the other factors gives us are derivative in a simplified form

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