Problem

Find the first, second, and third derivatives of …

Problem 9 Easy Difficulty

Find the first, second, and third derivatives of the given functions.
$$f(r)=r(4 r+9)^{3}$$

Answer

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Calculus 1 / AB

Basic Technical Mathematics with Calculus

Chapter 23

The Derivative

Section 9

Higher Derivatives

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Video Transcript

and this problem were asked to find e drg s Where are is equal to sq two minus two s squared plus three. So we know that this is just going to be ah lim. The limit is h closes. You're right. All derivatives are going to be limits. So where did each come in here? We need to plug in H s plus H to our function. So let's just write our of s plus h minus R of s over age. So that's the limit as H goes to zero of as plus age cubed minus two times as plus age squared plus three minus as cubed minus two. A squared plus three all over age. Let's expand everything here. So, Albert, it's a little bit messy. But don't worry, it'll it'LL work out pretty nicely in the end, let's expand out to smell it over here as cubed plus three as squared H plus three. Ask H squared plus age cubed minus two. A squared minus four A. C. H minus two A squared plus three minus as cubed minus two s squared plus three all over age. But now we get some nice cancellations here. So we have an s cubed minus as cube. So that cancels. Here we have a minus two as squared canceling with a minus two s squared Cancel an A plus three minus plus three. So those cancel But now we see that we're left with on ly h is here. So once we get rid of all these, you know, get rid of all these terms There's a tsh is life that we can now cancel The h is so h on the bottom is going to cancel So this age squared becomes h the power of one cancel With this age, execute becomes a squared, a squared becomes age, Age goes away So now here it's going to come up here We're left with the limit as h goes to zero So now after we've cancelled all of the ages three as squared plus three as H plus H squared minus for s minus two age So these are the terms that were left with So here was thes these five terms Now we can just set eight people to zero when we get three as squared plus three times as time zero zero squared minus for s minus two times zero. So our final answer is three s squared, minus four s. So here is the derivative. This is DRG s where are was rescued minus two s squared, plus three.

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

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