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# Find the derivative of the function.$F(t) = (3t - 1)^4 (2t + 1)^{-3}$

## $F^{\prime}(t)=\frac{12(3 t-1)^{3}(2 t+1)-6(3 t-1)^{4}}{(2 t+1)^{4}}$

Derivatives

Differentiation

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### Video Transcript

here we have a function that's a product. So this is the first factor and this is the second factor. So we want to find the derivative using the product rule and in the midst of using the product rule, we're going to be using the chain rule. So we're going to have the first times the derivative of the second plus two second times, the derivative of the first. So we start with the 1st 3 T minus one to the fourth power. Now we multiply by the derivative of the second. So we bring down the negative three and raised to t plus one to the negative fourth power. So that's the derivative of the outside function. Now we multiply by the derivative of the inside function and the derivative of To T plus one would be to. So what we have so far is the first times the derivative of the second. Let's keep going with the product rule. So we have plus the second to T plus one to the negative third times the derivative of the first, and will use the chain rule on that as well. So we bring down the four and we raise three T minus one to the third. That's the derivative of the outside. And then we multiply by three. The derivative of the inside. That's the derivative of three T minus one. Okay, now our goal is going to be to simplify this answer and their varying levels of simplifying that you might be required to do. It's just kind of depends on your instructors preferences. So what I'm going to do is I'm going to move the terms that have a negative exponents down to the denominators. And then I'm also going to multiply any Constance I have together like the negative three and the two almost apply those together and the four and the three on the other part. So here's what we have Now we have negative six times three T minus one to the fourth over to T plus one to the fourth. And then for the second part, we have 12 times three T minus one to the third over to T plus one to the third. Okay, so one thing we might want to do at this point is we might want to combine these into a single fraction with a common denominator And if that's what we're going to do, then we need to multiply the second fraction by to t plus one over itself. That way, the common denominator will be to t plus one to the fourth. So if we do that, we end up with all right, the denominator first to t plus one to the fourth. And the numerator is going to be now before, right, It noticed that this 1st 1 has a negative sign. So if we switch the order and put the second numerator first and the first numerator second, then we can just put a minus sign in between. So we have 12 times three t minus one, cubed times to t plus one minus six times three t minus one to the fourth all over the denominator and there is our derivative.

Oregon State University

Derivatives

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