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# Find the derivative of the function.$h(t) = (t +1)^{2/3} (2t^2 - 1)^3$

## $=\frac{2}{3}(t+1)^{-1 / 3}\left(2 t^{2}-1\right)^{2}\left(20 t^{2}+18 t-1\right)$

Derivatives

Differentiation

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

Okay, so here's h of tea and we're going to find its derivative. And this problem eyes probably about one part calculus and three parts algebra. We're going to pay real close attention to detail here, So we have a product T plus one to the 2/3 is a factor, and two t squared minus one. Cubed is a factor. So we're going to use the product rule, which is the first times a derivative of the second plus the second time's a derivative of the first. So let's get going on that. So the first would be t plus one to the 2/3 the derivative of the second. We would use the chain rule here, bring down the three and raised to t squared minus one to the second power that takes care of the derivative of the outside and then multiplied by the derivative of the inside. The derivative of two T squared minus one would be four teeth. Okay, so so far, we have the first times the derivative of the second. Now we need plus the second to T squared, minus one cubed times the derivative of the first. And so we're going to use the chain rule on that as well. Bring down the 2/3 and raise T plus one to the negative 1/3 and then multiply by the derivative of T plus one, which is just one, so I won't write that. Okay, so we have the second times, the derivative of the first. Okay, now we want to see this answer as to terms. This is one term, and this is one term, and we want to look for what we can factor out of both terms. Okay, So one thing we notice is that we have a to t squared minus one, squared in one, cubed in the other so we can factor out to t squared minus one squared. Now, is there anything else we can factor out? Well, take a look At T plus one to the 2/3 and t plus one to the negative. 1/3. When we're factoring out, we always factor out the Wiest, the smallest of those and the smallest of those would be t plus one to the negative 1/3. So how do you factor that out of t plus one to the 2/3? So let's do that off to the side T plus one to the 2/3 is equal to t plus one to the negative 1/3 times. What? That's really the question T plus one to the what? So these exponents have to add to 2/3. So the one in the box here would be 3/3 or one so that the two exponents could add to 2/3. Okay, so if we factor t plus one to the negative 1/3 out, we're going to be leaving t plus one. So let's go back and right what we have left in the first term, all of that. What we have left is T plus one times three times 40. Now what do we have left in the second term? So we have factored out to t squared minus one squared. So we still have a to t squared minus one, and we still have a 2/3. Okay, Now let's see what we can do to simplify this third factor. So we're keeping the first factor. As it is. We're keeping the second factor as it is. And now suppose we distribute so we would have 12 tee times the quantity t plus one. So that's going to be 12 t squared plus 12 t and suppose we distribute the 2/3 so that's going to be 4/3 T squared, minus 2/3. Okay, so we have some, like, terms we can combine. The 12 t squared and the 4/3 t squared. All right, so here's the step we were just looking at. We're combining these like terms. And now for answer, we have to t squared minus one squared times T plus one to the negative 1/3 times when we had those, like terms, we get 43rd T squared plus 12 t minus 2/3. Okay, so you may or may not be required to do all that simplifying. As a matter of fact, when I looked at the answer in the book, I saw something very similar, with slightly less simplifying. I think it was something like for the third term. It was something like 2/3 times a quantity T, um, to t squared minus one plus 12 t times a quantity T plus one. So same equivalent last term. Compared to what I have just not simplified as much

Derivatives

Differentiation

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