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Find $f^{\prime \prime}(x)$.$$f(x)=\left(x^{2}+3 x\right)^{7}$$

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$2\left(x^{2}+3 x\right)^{5}\left(91 x^{2}+273 x+189\right)$

Calculus 1 / AB

Chapter 1

Differentiation

Section 8

Higher-Order Derivatives

Derivatives

Missouri State University

Campbell University

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

00:22

Find $f^{\prime}(x)$.$…

01:35

Find $f^{\prime}(x)$$$…

01:47

We need to get right into finding a second derivative of this problem because we start with the chain rule for the first derivative. But then we need to move on to the chain and the product for the second directive. So let's start with the first derivative because you're married to find the secondary evidence. A derivative of the derivative. This is changeable where you bring the seven in front and now it's X squared. Plus three X is to the six power. But then you need to multiply by the derivative of that function, the inner part, which is two X plus three. Well, now and I tried toe emphasize it that we have a product in here. So the first year of the logistic, the changeable Well, now we have a product, which means that the second derivative is a product and a changeable Well, if I take the derivative of the first piece, the left side of the product you know you bring the six in front at the B 42 x squared plus three X would be now taken to the fifth power. You tracked one from your exponents and you take the derivative of the inside, which is two X plus three. Well, remember, we leave the right side alone in the product rule so you could just write, like instead of writing a second time. There's two of them now, so squared. And then plus, now you leave the first part alone at seven times X squared plus three x to the six power notice. I didn't write the seven here because the derivative of this piece derivative of two X plus three is too. So if I were to multiply by that seven, I would get 14. Um, yes. So this is your answer. A. A different math book might factor things out. I don't see any benefit of doing that. Like you could factor out five of these X squared plus three X didn't even factor out of seven or 14 between the two terms. But there's really no benefit to doing that. You have the right answer right here.

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