find-the-derivatives-of-the-given-functions-y4-etlefte2-t-etright-3

Question

Find the derivatives of the given functions.
$$y=4 e^{t}\left(e^{2 t}-e^{t}\right)$$

Tyler Moulton · Accepted Answer

Uh huh. We want to find Dy DT for the function Y equals for each of the tee times either to t minus E to the T. We&#x27;re gonna rely on derivative shortcuts to solve this. Specifically, rule 0 to 3. Here we&#x27;re rule Zero states GDX BTU is B D u L n B D u d x G D X. Either the U. Is either the U d u d X rules one through three. Other powerful product rule in general, respectively. Before we use any of these will to solve, let&#x27;s rewrite why? In a form that&#x27;s more easily differentiable. So we have Y equals for each of the T. Either to t minus for each of the t V, the T. Or more simply for each of the three t minus for each of the two T. Thus we can simply apply rule zero to solve this. So we have Dy DT is equal to 14 of the three t times the u d x three minus four each of the two t time dutt. To simplifying, we have final solution. DY DT is equal to 12 each of the three t minus eight each of the two t.

Bobby Barnes · Answer

we want to find the derivative of the team is able to t squared minus one over T squared plus team minus two. So you could just start by using the caution war to take the derivative of this. But before we do that, let&#x27;s see if we can back this to possibly supply a little bit. So our numerator will too t plus one t minus one difference of squares and our denominator here. Well, we ask yourselves what number can multiply a negative to or what? Seven numbers won&#x27;t find negative too. But after one Principe&#x27;s t cost you over T minus one man, you might notice that he&#x27;s t minus ones or counts. And now we have t plus one over t plus two, which significantly simplifies what we&#x27;re trying to do here. And this one thing you need to make sure to write up onto the side that t cannot equal to what So I&#x27;ll take the drivel about t plus one of the teapots too. So, like I was saying, this is crucial. And so just to remind you of the suits so quotient rule. So the way I remember is the following. So ddx of some high function over some No function. So it&#x27;s slow. D Hi minus. Hi. Hello. All over the square of what&#x27;s below. So in this case, our highest people&#x27;s wanna Arlo is t plus two so we can Hey, t x t x there&#x27;s gonna bea t in this case dd t oh t cost one over t plus two. Well, this is going to be low. So maybe it&#x27;Ll be a good idea if I read these in different colors. Oh, all right. I like the new male Blue will be team plus one on the denominator green tea plus two so we can take low first. So t cross too. Team of what we have in the numerator t plus one minus what we have in the new mayor t plus one time to drive. They were respected T t too all over what we have in the denominator squared two plus two square. So the derivative of t plus one and keep us too Well, both of these are in the form out a derivative of t course from constant C, so we could go ahead and use some outstanding property to first break us. You team plus T c. The derivative of a constant is zero and a derivative of T to the first power will just be one. Since we could move the one out front using power rule and their team, you should try that. Get tea to zero and then teach zero is just what? So each of these derivatives here and become one So have I t plus two times one minus t plus one times one over t lost too square No. Now T minus t is going to counsel out and then two minus one is one So we can rewrite this as one over t plus two squared And then just recalling that this function should not be defied at T minus one and likewise are derivative. Should also be not to find that tea is equal to what So this is our derivative and t does not equal one is just one of the other values. This derivative is not supposed

Nicholas Barvinok · Answer

we have the function e to the minus two tea Co sign a 40 and we want to take its derivative. So to take its derivative, we have a product to functions. So we will be using the product rule for derivatives That states that the derivative of F times G is F prime G plus f g prime. So in this case we have the derivative of e T minus two t. So we use the chain rule. The outer function is e to the input of that function the derivative that is e to the input of that function. Then times the derivative of that input which is minus two. Don&#x27;t put the minus two out here and then times the second function coastline of 40 plus the first function e to the minus two tea times the derivative off the second function CO sign of 14 Here again, we use the chain rule. We have an inter function 40 in an outer function co sign of some input. So by the chain rule, we take the derivative of the outer function first co sign of input. The derivative of that is minus sign of the same input Then we multiply by the derivative of the inter function for tea, the derivative that is four. And we can rewrite this a bit as we can. Factor out a minus two e to the minus two tea and we get co sign for 13 plus two. Sign for tea and we&#x27;re done.

Linh Vu · Answer

in discussing different functions. Quiet encoded each of the barrel for X and it wasn&#x27;t asked to find that the remedy under function. So why prime we, coach, you were recording done for any function exponential function h it about off the huge X and then we want to find a derivative and we go to each it about you to the X times. Good Do Graham on the axe. So using this formula here, we can identify This one will be the judge of that X and therefore my family. We&#x27;re gonna aged off x times with a far X derivative so for exterior. But if you go too far and temperament agent for X definitely can write, can&#x27;t answer echoed uniform times agent for X and this will be the wiper and we&#x27;re looking for here.

Alex Roush · Answer

Okay, so we have to take the derivative of this function. Now, if you have not learned implicit differentiation or the chain rule, you do have to use the quotient rule on this. Otherwise, you could actually just bring this guy up and rewrite. This entire function is to to the 4/3 eat in the negative t and you can use the product rule and you would know how to differentiate this. But I&#x27;m assuming up to now that you don&#x27;t know how to differentiate anything other than just eat of the tia eso we&#x27;re just gonna use the sociable. You&#x27;re so s prime of tea. I would therefore be lo de Hi. I&#x27;m just gonna go and take the derivative this It&#x27;s gonna be you 4/3 t year to the 4/3 minus one. Why? Because the power role says that we would move this down, is a coefficient and then subtract one minus high d low. It&#x27;s the derivative of the exponential is just the exponential divided by swearing about. I could doctor and eat of the tea out and I&#x27;m left with what, 4/3 t. What&#x27;s the exponents? Well, 4/3 minus one is 4/3 minus 3/3 which is 1/3. Okay, so I factored this guy out, and I also subtract a t to the 4/3 right now I&#x27;m dividing by each of the t squared. But you know, the T squared is either the T year tens eat of the tear, which is, you know, the T plus tty. If you wanted to factor a t to the 1/3 out, some instructors might like this. But my sake, this is good enough other than you can cancel on each of the t. O. Okay, so I&#x27;m actually going to I almost didn&#x27;t catch that. Okay, so if I divide, I guess multiply those sides by an eat in the negative t I can do because it&#x27;s the same thing is multiplying by one. I&#x27;m not changing the expression at all. Then these cancel out to form one, and then I have either the two tier times e to the negative T, which ends up being just one of a team. And then, if you really wanted to factor the sky out, you could factor a t to the 1/3 out, and then you&#x27;d be left with 4/3 minus minus What? Minus t right. This could be your correct answer.

Hayden Woerner · Answer

all right are derivative here, the derivative of e sorry for each of the two x squared. We leave the four cause it&#x27;s a coefficient. We multiply by the derivative of E to the two x squared, which is e to the two X squared times, the derivative of two X squared, which is four X. Then we mult. We subtract from that dirt the derivative of four, which is zero. So our final answer here we can put the four and the four together to be 16 x e to the two x squared.

Problem

Find the derivatives of the given functions. $$y=…

Problem 7 Easy Difficulty

Find the derivatives of the given functions.
$$y=4 e^{t}\left(e^{2 t}-e^{t}\right)$$

Answer

$\frac{d y}{d t}=12 e^{3 t}-8 c^{2 t}$

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Caleb E.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

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$\frac{d y}{d t}=12 e^{3 t}-8 c^{2 t}$

Basic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Caleb E.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Grace H.

Anna Marie V.

Caleb E.

Samuel H.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus