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Find $d s / d t$ if $s=\left(3 t^{2}+1\right)^{2}\left(t^{3}-6 t+1\right)^{7}$.

$$3\left(3 t^{2}+1\right)\left(t^{3}-6 t+1\right)^{6}\left(25 t^{4}-59 t^{2}+4 t-14\right)$$

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 6

The Chain Rule

Derivatives

Missouri State University

Oregon State University

Harvey Mudd College

University of Nottingham

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.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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02:16

if f(t) = (t^2 + 6t +7)(3t…

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Find $d s / d t$$$…

01:59

Find $\frac{d}{d t}[\mathb…

00:42

00:51

we have to find the derivative of this function and we're going to implore thee, product rule and the general. So we're looking at three t squared plus one and that quantity is squared. And then times there's your product rule. We have teak minus 60 plus one all to the seventh, and that's gonna be a chain room. So, as I mentioned, find your product rule. The reason why that's important is because to the left, when we start to do the derivatives D S. D. T. 10, what we have to do is the change Roll to the left, which is bringing that to in front. You leave three t squared, plus one alone. It's now to the first power where you have to write that and then times the derivative of the inside, which would be 16. But then you also have to leave the right side of that alone so that t cubed minus 60 plus one to the seventh power is left alone and then plus, now you leave the left side alone. So that's the quantity of three t squared plus one squared and then times the derivative of the right side, which is the changeable where you bring the seven in front and then you leave it cubed, minus 60 plus one alone. It's now to the six power times, the derivative of the inside to be three t squared 96. Now some teachers will let you leave your answer like that long enough as this e. It's difficult enough as it is otherwise. What you can do is you can identify things their shared between both terms. For instance, they each share at least 13 T squared, plus one what else is shared. They share six of these T Q minus 60 plus one eso you could factor that helped Q minus 60 plus one to the six power, and then you'll be left with a quantity, and it would take a lot of math to factor out. Um, I think what they also do is they factored out of three. Here, um, that you can combine with the three C three out of the six eso after you do a lot of algebra. This would take a lot of algebra and probably be a waste of time. If you want my opinion to get 25 t to the fourth minus 50 90 squared. Probably thinking that that this is a waste of time and I would agree plus 40 and then minus 14. And I hope your teacher does not make you do this algebra because this is all algebra. It's no longer calculus when you get there. Eso I would have stopped at my red and blue equation that already circled. Otherwise, you can simplify to this answer down here.

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