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Find the derivative of the function. $ J(\theta)…

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Problem 29 Medium Difficulty

Find the derivative of the function.
$ H(r) = \frac {(r^2 - 1)^3}{(2r + 1)^5} $

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April 16, 2020

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Video Transcript

here we have a function that's a quotient. So in finding the derivative, we're going to use the quotient rule. And in the sun, inside of the quotient, we have some composite functions, so we'll be using the chain rule as we go. So h prime of our would be according to the quotient rule. We have the bottom to r plus one to the fifth times the derivative of the top. So here we're going to stop and use the chain rule. We bring down the three and we take our squared minus one to the second. And then we multiply by the derivative of the inside, which would be to our so So far, we have the bottom times, a derivative of the top. Now we need minus the top minus R squared minus one, cubed times the derivative of the bottom on the derivative of the bottom. We're going to need to use the chain rule. So bring down the five and raised to R plus one to the fourth and then multiply by the derivative of the inside, which is to and that's all over the bottom squared. So that's all over to R plus one to the 10th. Okay, The rest of it is all about simplifying. What are we going to do to simplify this? So let's look at this whole first part as the first term and this whole second part as the second term. And let's see if they have anything in common that we can factor out. So they both have a factor of two. They both have a factor of to R plus one to the fourth. This one has an extra that we're going to leave behind. And this one has a two r plus one to the fourth. So let's factor that out. And they both have a factor of R squared minus one squared. The 2nd 1 has an extra that we're going to leave behind. It will factor R squared minus one squared. Now, what did we leave behind? So in the first part, we left one of the two our plus ones and we left the three and the are So we have three are times a quantity to R plus one. What did we leave behind in the 2nd 1? Well, we still have the minus, and then we have one of the R squared, minus ones that was left behind. And we have the five. So we have minus five times R squared minus one. Okay. And that's all over to R plus one to the tent. So what we can do now is reduce our to our plus one to the fourth over to R plus one to the 10th. And that would just leave us with two r plus one to the sixth on the bottom. And then the next thing we can do is distribute and simplify this quantity on the brackets. Okay, We're going to grab a little extra space to simplify the quantity in the brackets. So we'll distribute. The three are and we get six r squared plus three are and we'll distribute the minus five when we get minus five r squared plus five and will combine like terms. And we have our squared plus three are plus five. Okay, so let's put that back into our answer. So we have two times R squared minus one squared times are squared plus three R plus five. That quantity. We just simplified over to R plus one to the sixth

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Find the derivative of the function. $ J(\theta)…

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Watch More Solved Questions in Chapter 3

Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Samuel Hannah

Michael Jacobsen

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Samuel Hannah

Michael Jacobsen

James StewartCalculus: Early Transcendentals

Catherine Ross

Caleb Elmore

Samuel Hannah

Michael Jacobsen

James Stewart

Calculus: Early Transcendentals